Quantity  Unit  Unit symbol 
Length  meter  m 
Mass (or weight)  gram  kg 
Volume  litre  L 
Time  second  s 
Temperature  degree (Celsius)  °C 
Prefix  Symbol (abbreviation)  Power of 10  Multiple value  Example 
mega  M  10^{6}  1,000,000  1 Mm = 1,000,000 m 
kilo  k  10^{3}  1,000  1 km = 1,000 m 
hecto  h  10^{2}  100  1 hm = 100 m 
deka  da  10^{1}  10  1 dam = 10 m 
meter/gram/liter  1  
deci  d  10^{1}  0.1  1 m = 10 dm 
centi  c  10^{2}  0.01  1 m = 100 cm 
milli  m  10^{3}  0.001  1 m = 1,000 mm 
micro  µ  10^{6}  0.000 001  1 m = 1,000,000 µm 
Prefix  Length (m  meter)  Weight (g  gram)  Liquid volume (L  liter) 
mega (M)  Mm (Megameter)  Mg (Megagram)  ML (Megaliter) 
kilo (k)  km (Kilometer)  kg (Kilogram)  kL (Kiloliter) 
hecto (h)  hm (hectometer)  hg (hectogram)  hL (hectoliter) 
deka (da)  dam (dekameter)  dag (dekagram)  daL (dekaliter) 
meter/gram/liter  m (meter)  g (gram)  L (liter) 
deci (d)  dm (decimeter)  dg (decigram)  dL (deciliter) 
centi (c)  cm (centimeter)  cg (centigram)  cL (centiliter) 
milli (m)  mm (millimeter)  mg (milligram)  mL (milliliter) 
micro (µ)  µm (micrometer)  µg (microgram)  µL (microliter) 
 Convert a smaller unit to a larger unit: move the decimal point to the left.
 Convert a larger unit to a smaller unit: move the decimal point to the right.
Example 1.3.1
Count places from mm to m: 3 places left
Convert a smaller unit (mm) to a larger (m) unit: move the decimal point to the left.
326. mm = 0.326 m  Move the decimal point three places to the left (326 = 326.). 
Example 1.3.2
Count places from hg to g: 2 places right
Convert a larger unit (hg) to a smaller (g) unit: move the decimal point to the right.
4.765 hg = 476.5 g  Move the decimal point two places to the right. 
Example 1.3.3
$= \frac{1200 kg}{1000}$
$= 1.2 kg$
Example 1.3.4
“mm” is the desired unit.
$= \frac{(30)(10) mm}{1}$
$= 300 mm$
Example 1.3.5
3 m  3000 mm  1 m = 1,000 mm  
– 2000 mm  >  – 2000 mm  
1000 mm 
25 kg  25000 g  1 kg = 1000 g  
+ 4 g  >  + 4 g  
25004 g 
Example 1.3.6
1) 16cm^{3} = ( ? ) g  
16cm^{3} = 16 g  1 cm^{3} = 1 g 
2) 9 L = ( ? ) cm^{3}  
9 L = 9000 mL  1 L = 1,000 mL 
= 9000 cm^{3}  1 mL = 1 cm^{3} 
3) 35 cm^{3} = (?) cL  
35cm^{3} = 35 mL  1 cm^{3} = 1 mL 
= 3.5 cL  move 1 decimal place left. 
4) 450 kg = (?) L  
450 kg = 450,000 g  1 kg = 1,000 g 
= 450,000 mL  1 g = 1 mL 
= 450 L  1 L = 1,000 mL 
Example 1.3.7
V = w l h = (8m) (10m) (2m) = 160 m^{3}  160 m^{3} = ( ? ) kL 
160m^{3} = 160,000,000 cm^{3}  1 m = 100 cm, 3 × 2 = 6, move 6 places right for volume. 
160,000,000 cm^{3} = 160,000,000 mL  1 mL = 1 cm^{3} 
160,000,000 mL = 160 kL  1 kL = 1,000,000 mL 
160 m^{3} = 160 kL  The swimming pool will hold 160 kL of water. 
a. 439 mm = ( ? ) m
b. 3 mL = ( ? ) kL
c. 7230 g = ( ? ) kg
d. 52 cm = ( ? ) mm
3. Combine:a. 7 m – 3000 mm = ( ? ) mm
b. 63 kg + 6 g = ( ? ) g
4. Complete:a. 38 cm^{3} = ( ) g
b. 5 L = ( ) cm^{3}
c. 18 L of water has a volume of ___________ cm^{3} .
d. A water tank measures 45 cm by 35 cm by 25 cm. How many kiloliters of water will it hold?
Answers
Algebraic term  Description  Example 
Algebraic expression  A mathematical phrase that contains numbers, variables (letters), and arithmetic operations (+, – , ×, ÷, etc.). 
3x – 4 5a^{2} – b + 3 12y^{3} + 7y^{2} – 5y + $\frac{2}{3}$ 
Constant  A number on its own.  2y + 5 constant: 5 
Coefficient  The number in front of a variable. 
9x^{2} coefficient: 9 x coefficient: 1 (x = 1 · x) 
Term  A term can be a constant, a variable, or the product of a number and variable. (Terms are separated by a plus or minus sign.) 
2x^{3} + 7x^{2} – 9y – 8 Terms: 2x^{3}, 7x^{2},  9y, 8 
Like terms  The terms that have the same variables and exponents (differ only in their coefficients). 
2x and 7x 4y^{2} and 9y^{2} 0.5pq^{2} and $\frac{2}{3}$pq^{2} 
Example: 7x + 9 , 9t^{2} – 2t , 0.3y + $\frac{1}{3}$
Example  Like or unlike terms 
7y and 9y  Like terms 
6a^{2}, 32a^{2}, and a^{2}  Like terms 
0.3 x^{2}y and 48x^{2}y  Like terms 
$\frac{2}{7}$u^{2}v^{3} and u^{2}v^{3}  Like terms 
8y and 78x  Unlike terms 
6m^{3} and 9m^{2}  Unlike terms 
9u^{3}w^{2} and 9w^{3}u^{2}  Unlike terms 
Example 1.1.1
1) 3a + 7b – 9a + 15b = (3a – 9a) + (7b + 15b)  Regroup like terms. 
= 6a + 22b  Combine like terms. 
2) 2y^{2} – 4x + 3x – 5y^{2} = (2y^{2} – 5y^{2}) + (4x + 3x)  Regroup like terms. 
= 3y^{2} – 1x  Combine like terms. 
= 3y^{2} – x 
3) 8xy^{2} – x^{2}y + 4x^{2}y – 6xy^{2}  
= 8xy^{2} – x^{2}y + 4x^{2}y – 6xy^{2}  Or underline like terms without regrouping. 
= 2xy^{2} + 3x^{2}y  Combine like terms. 
4) 2(2m + 3n) + 3(m – 4n) = 4m + 6n + 3m – 12n  Distributive property. 
= 7m – 6n  Combine like terms. 
Algebraic expression  Remove parentheses  Example 
(ax + b)  ax + b  (5x + 2) = 5x + 2 
(ax – b)  ax – b  (9y – 4) = 9y – 4 
 (ax + b)  ax – b   ($\frac{3}{4}$x + 7) =  $\frac{3}{4}$x – 7 
 (ax – b)  ax + b   (0.5b – 2.4) = 0.5b + 2.4 
Example 1.1.2
1) 9x^{2} + 7 – (2x^{2} – 2) = 9x^{2} + 7 – 2x^{2} + 2  Remove parentheses. 
= 7x^{2} + 9  Combine like terms. 
2) (8y + 5z) – 4(y – 7z) = 8y + 5z – 4y + 28z  Remove parentheses. 
= 12y + 33z  Combine like terms. 
3)  (3a^{2} + 4a – 4) + 3(4a^{2} – 6a + 7)  Remove parentheses. 
=  3a^{2} – 4a + 4 + 12a^{2} – 18a + 21  Distributive property. 
= 9a^{2} – 22a + 25  Combine like terms. 
4) 5(u^{2} – 3u) + 3(2u – 4) – (5 – 3u + 4u^{2})  Distributive property. 
= 5u^{2} + 15u + 6u – 12 – 5 + 3u – 4u^{2}  Remove parentheses. 
= 9u^{2} + 24u – 17  Combine like terms. 
Example 1.1.3
1) 3x^{3} (5x^{2} – 2x) = (3x^{3}) (5x^{2}) – (3x^{3}) (2x)  Distributive property: a (b + c) = ab + ac 
= (3 ∙ 5) (x^{3} x^{2}) – (3 ∙ 2) (x^{3} x^{1})  Regroup x = x^{1} 
= 15 (x^{3+2}) – 6 (x^{3+1})  Multiply the coefficients & add the exponents. 
= 15x^{5} – 6x^{4}  a^{m} ∙ a^{n} = a^{m+n} 
2) 5ab^{2} (2a^{2}b +ab^{2} – a)  Distribute. 
= (5ab^{2}) (2a^{2}b) + (5ab^{2}) (ab^{2}) + (5ab^{2}) (a)  Multiply the coefficients and add exponents. 
= (5 ∙ 2) (a^{1+2} b^{2+1}) + (5a^{1+1} b^{2+2}) – (5a^{1+1}b^{2})  b = b^{1} , a = a^{1} 
= 10a^{3}b^{3} + 5a^{2}b^{4} – 5a^{2}b^{2}  a^{m} ∙ a^{n} = a^{m+n} 
Example 1.1.4
$\frac{12x^2+4x2}{4x}$
Steps Solution
(a + b) (c + d) = ac + ad + bc + bd F O I L  Example 
F  First terms  first term × first term (a + b) (c + d)  (x + 5) (x + 4) 
O  Outer terms  outside term × outside term (a + b) (c + d)  (x + 5) (x + 4) 
I  Inner terms  inside term × inside term (a + b) (c + d)  (x + 5) (x + 4) 
L  Last terms  last term × last term (a + b) (c + d)  (x + 5) (x + 4) 
FOIL method  Example 
(a + b) (c + d) = ac + ad + bc + bd  (x + 5) (x + 4) = x ∙ x + x ∙ 4 + 5x + 5 ∙ 4 = x^{2} + 9x + 20 
F O I L  F O I L 
Example 1.1.5
1) $(2x + 3)(5x  6)=2x \cdot 5x+2x( 6)+3 \cdot 5x+3(6)$ F O I L  The FOIL method. 
$=10x^212x+15x18$  a^{n} a^{m} = a^{n+ m} 
$=10x^2+2x18$  Combine like terms. 
2) $(3r – t)(5r+t^2)=3r \cdot 5r+3r \cdot t^2t \cdot 5rt \cdot t^2$  FOIL 
$=15r^2+3rt^25rtt^3$  a^{n} a^{m} = a^{n+m} 
$=18r^2+5rtt^3$  Combine like terms. 
3) $(xy^2+y)(2x^2y+x)=xy^2 \cdot 2x^2y+xy^2 \cdot x+y \cdot 2x^2y+yx$  FOIL 
$=2x^3y^3+x^2y^2+2x^2y^2+xy$  a^{n} a^{m} = a^{n+m} 
$=2x^3y^3+3x^2y^2+xy$  Combine like terms. 
4) $(a\frac{1}{3})(a\frac{1}{3})=a^2\frac{1}{3}a\frac{1}{3}a+(\frac{1}{3})(\frac{1}{3})$  FOIL 
$=a^2\frac{2}{3}a+\frac{1}{9}$  Combine like terms. 
a. 5x^{3} 8x^{2} + 2x
b. –$\frac{2}{3}$y^{4} + 9a^{2} + a – 1
2. Combine like terms:a. 7x + 10y – 8x + 9y
b. 12a^{2} – 33b + 2b – 6a^{2}
c. 13n + 5(6n – m^{2}) + 7(2m^{2} + 3n)
3. Simplify:a. 15a^{2} + 9 – (5a^{2} – 4)
b. (13x + 9y) – 6(x – 5y)
c. 5(ab – 2xy) – 6(2ab + 3xy)
d. (5y – 7) (8y + 9)
e. (7r – 2t) (3r + 4t^{2})
f. (x – $\frac{1}{3}$) (x – $\frac{2}{3}$)
Answers
Example 1.2.1
1) 2 : 4x – 3 = 5  4 ∙ 2 – 3 $\overset{?}{=}$ 5  5 $\overset{\checkmark}{=}$ 5  Yes  Replace x with 2. 
2) 15 : $\frac{3}{15}$y = 3  $\frac{3}{15}(15) \overset{?}{=}$ 3  3 $\overset{\checkmark}{=}$ 3  Yes  Replace y with 15. 
3) $\bf\frac{1}{2}$ : 8t = 3  8 ($\frac{1}{2}$) $\overset{?}{=}$ 3  4 ≠ 3  No  Cannot replace t with $\frac{1}{2}$. 
Properties  Equality  Example 
Addition property of equality  A = B A + C = B + C  Solve $x6=3$ $x\bcancel{6}+\bcancel{\bf6}=3+\bf6$ x = 9 
Subtraction property of equality  A = B A – C = B – C  Solve $y+5=8$ $y+\bcancel{5}\bcancel{\bf5}=8\bf5$ y = 13 
Multiplication property of equality  A = B A · C = B · C  Solve $\frac{m}{9}=2$ $\bcancel{\bf9} \cdot \frac{m}{\bcancel{9}}=2 \cdot \bf9$ m = 18 
Division property of equality  A = B $\frac{A}{C}=\frac{B}{C}$ (C ≠ 0)  Solve $3n=15$ $\frac{\bcancel{3}n}{\bcancel{\bf3}}=\frac{15}{\bf3}$ n = 5 
Example 1.2.2
1) $9 + x = 5$  $\bcancel{9} + x+ \bcancel{\bf9} = 5 + \bf9$  Property of addition. 
$x = 14$  
Check:  $9 + {\bf14} \overset{?}{=} 5$ $5 \overset{\checkmark}{=} 5$  Replace x with 14. 
2) $t+\frac{2}{5}=\frac{1}{5}$  $y+\frac{2}{5}\frac{\bf2}{\bf5}=\frac{1}{5}\frac{\bf2}{\bf5}$  Property of subtraction. 
$y=\frac{3}{5}$ 
3) $\frac{1}{6}x = 7$  ${\bf6} \cdot \frac{1}{6}x=7(\bf6)$  Property of multiplication. 
$x = 42$ 
4) $5x = 30$  $\frac{5x}{\bf5}=\frac{30}{\bf5}$  Property of division. 
$x = 9$ 

Steps  Example 
Solve $\bf\frac{1}{5}(y+10)=3y\frac{9}{5}y$  

$\bcancel{5}\cdot \frac{1}{\bcancel5}(y+10)={\bf5}(3y){\bf\bcancel5}(\frac{9}{\bcancel5}y)$  Multiply each term by 5. 

$y+10=15y9y$  

$y+10=6y$  

$y+\bcancel{10}\bcancel{\bf10}=6y\bf10$ $y=6y\bf10$ $y{\bf6y}=\bcancel{6y}10\bcancel{\bf6y}$ 
Subtract 10 from both sides. Subtract 6y from both sides. 

$5y=10$ $y=\frac{10}{5}$ $y=2$  Divide both sides by 5. 

$\frac{1}{5}(2+10)=3 \cdot 2\frac{9}{5} \cdot 2$ $\bcancel{5}\frac{1}{\bcancel5}(2+10)=5 \cdot 3 \cdot 2\bcancel{5} \cdot \frac{9}{\bcancel5} \cdot 2$ $(2+10)=3018$ $12=12$  Replace y with 2. Multiply each term by 5. LS = RS (correct) 
Steps  Example 
Solve $\bf0.34x  0.12 = 4.26x$  

${\bf100}(0.34x)  {\bf100}(0.12) = {\bf100}(4.26x)$  The largest number of decimal place is two. 

$34x12 = 426x$ $34x + 426x = 12$ $460x = 12$ 
Add 12 to both sides. Add 426x to both sides. 

$x \approx 0.026$ 
Example 1.2.3
Solve 0.4y + 0.08 = 0.016  The largest number of decimal place is three. 
1000(0.4y) + 1000(0.08) = 1000(0.016)  Multiply each term by 1000. 
400y + 80 = 16  Combine like terms. 
400y = 64  Divide both sides by 400. 
y =  0.16 
Steps  Example 
Solve $\bf\frac{t}{3}+\frac{3}{4}=\frac{t}{2}\frac{1}{3}$  

${\bf12} \cdot \frac{t}{3}+{\bf12} \cdot \frac{3}{4}={\bf12}(\frac{t}{2}){\bf12} \cdot \frac{1}{3}$  

$4t+9=6t4$ $10t=13$ 
Add 6t to both sides. Subtract 9 from both sides. 

$t=\frac{13}{10}=1\frac{3}{10}$  Divide both sides by 10. 
Addition (+)  Subtraction ()  Multiplication (×)  Division (÷)  Equals to (=) 
add  subtract  times  divided by  equals 
sum (of)  difference  product  quotient  is 
plus  take away  multiplied by  over  was 
total (of)  minus  double  split up  are 
altogether  less (than)  twice  fit into  were 
increased by  decreased by  triple  per  amounts to 
gain (of)  loss (of)  of  each  totals 
combined  (amount) left  how much (total)  goes into  results in 
in all  savings  how many  as much as  the same as 
greater than  withdraw  out of  gives  
complete  reduced by  ratio/rate  yields  
together  fewer (than)  percent  
more (than)  how much more  share  
additional  how long  average 
Example 1.2.4
1) Edward drove from Prince George to Williams Lake (235 km), then to Cache Creek (203 km) and finally to Vancouver (390 km). How many kilometers in total did Edward drive?
235km + 203 km + 390 km = 828 km  The key word: total (+) 
2) Emma had \$150 in her purse on Friday. She bought a pizza for \$15, and a pair of shoes for \$35. How much money does she have left?
\$150 – 15 – 35 = \$100  The key word: left (–) 
3) Lucy received \$950 per month of rent from Mark for the months September to November. How much rent in total did she receive?
\$950 $\cdot$ 3 = \$2850  The key word: how much total (×) 
4) Julia is going to buy a \$7500 used car from her uncle. She promises to pay \$500 per month. In how many months she can pay it off?
\$7500 ÷ \$500 = 15 month  The key word: per (÷) 
 Organize the facts given from the problem (create a table or diagram if it will make the problem clearer)  Identify and label the unknown quantity (let x = unknown).  Convert words into mathematical symbols, and determine the operation – write an equation (looking for ‘key’ or ‘clue’ words).  Estimate and solve the equation and find the solution(s).  Check and state the answer. (Check the solution to the equation and check it back into the problem – is it logical?) 
Example 1.2.5
5 socks  \$4.35 each 
Sales tax  \$2.15 
Money left  \$5.15 

\$4.35 × 5 

(\$4.35 × 5) + \$2.15 

x = [(\$4.35 × 5) + \$2.15] + \$5.15 

x = [(\$4 × 5) + \$2] + \$5 
= \$27  

x = [(\$4.35 × 5) + \$2.15] + \$5.15 
= \$29.05 
\$29.05  [(\$4.35 × 5) + \$2.15] $\overset{?}{=}$ \$5.15  
\$29.05 – \$23.9 $\overset{\checkmark}{=}$ \$5.15  Correct! 
Example 1.2.6
James had  96 toys 
The total number of toys sold  13 + 32 + 21 + 14 + 7 
The toys not sold  96 – the total number of toys sold 

Let x = percentage of the toys were not sold 

13 + 32 + 21 + 14 + 7 = 87 

96 – 87 = 9 

x = $\frac{Toys\,not\,sold}{Total\,number\,of\,toys}$ = $\frac{9}{96} \approx$ 0.094 = 9.4% 
Example 1.2.7
60 L × $\frac{1}{2}$ = 30 L  Robert has 30 liters gas in his car. 
$\frac{7L}{100 km} = \frac{x}{390 km}$  Proportion: $\frac{a}{b} = \frac{c}{d}$ 
(x)(100km) = (7 L) (390 km)  Cross multiply and solve for x. 
x = $\frac{(7 L)(390 km)}{100 km}$ = 27.3 L  Robert needs 27.3 liters gas to get to Vancouver. 
a. x – 7 = 12
b. y + $\frac{3}{8}$ = $\frac{5}{8}$
c. $\frac{3x}{2}$ = $\frac{9}{16}$
d. 14t + 5 = 8
e. 7(x – 3) + 3x – 5 = 2(5 – 4x)
f. $\frac{1}{7}$(y + 12) = 4y – $\frac{3}{7}$y
g. 0.5t + 0.05 = 0.025
h. $\frac{x}{4}+\frac{2}{5}=\frac{x}{2}\frac{1}{5}$
2. Write an expression for each of the following:a. Susan has \$375 in her checking account. If she makes a deposit of y dollars, how much in total will she have in her account?
b. Mark weighs 175 pounds. If he loses y pounds, how much will he weigh?
c. A piece of wire 45 meters long was cut in two pieces and one piece is w meters long. How long is the other piece?
d. Emily made 4 dozen muffins. If it cost her x dollars, what was her cost per dozen muffins? What was her cost per muffin?
Answers
The ratio of a and b is: a to b or a : b or $\frac{a}{b}$
5 to 9 or 5 : 9 or $\frac{5}{9}$
 Write the ratio in a fractional form.
 Simplify and drop the units if given (as they cancel each other out).
Example: $4 : 28 = \frac{\bcancel{4}}{\bcancel{28}}=\frac{1}{7}$
$\div4$
Example: 0.75 meters to 0.25 meters $\frac{0.75m}{0.25m}=\frac{\bcancel{75}}{\bcancel{25}}=\frac{3}{1}=3$
$\times100$ $\div25$
$\frac{2\;teachers}{83\;students}$ , $\frac{24\;dollars}{3\;hours}$ , $\frac{85\;miles}{2\;hours}$
÷80
 Miles (or kilometres) per hour (or minute).
 Cost (dollars/cents) per item or quantity.
 Earnings (dollars) per hour (or week).
Example 1.4.1

4 L milk  2 L milk 
\$4.38  \$ x = ? 

$\frac{4\;L}{\\$4.38}=\frac{2\;L}{\\$x}$  

$(4)(x)=(2)(4.38)$  

$\frac{\bcancel{4}x}{\bcancel{4}}=\frac{2(4.38)}{4}$  Divide both sides by 4. 
$x=\frac{(2)(4.38)}{4}=2.19$  
2 liters of milk cost \$2.19.  

$\frac{4\;L}{\\$4.38}=\frac{2\;L}{\\$2.19}$  Replace x with 2.19. 
(4) (2.19) = (2) (4.38)  
8.76 = 8.76  Correct! 
Example 1.4.2

Tom’s height = 1.75 m  Building’s height (x) = ? 
Tom’s shadow = 1.09 m  Building’s shadow = 10m 

$\frac{1.75m}{1.09m}=\frac{xm}{10m}$  

$(1.75)(10)=(1.09)(x)$  

$x=\frac{(1.75)(10)}{1.09}=\frac{(\bcancel{1.09})x}{\bcancel{1.09}}$  Divide both sides by 1.09. 
$x=\frac{(1.75)(10)}{1.09}\approx16.055$  
The building’s height is 16.055m.  

$\frac{1.75m}{1.09m}=\frac{16.055m}{10m}$  Replace x with 16.055. 
$(1.75) (10) = (16.055) (1.09)$  
$17.5 = 17.5$  Correct! 
Example 1.4.3

$\frac{15\;mL}{180\;mL}=\frac{x\;mL}{230\;mL}$  $\frac{15\;mL\;medicine}{180\;mL\;water}=\frac{x\;mL\;medicine}{180\;mL\;water}$ 

$(15)(230)=(180)(x)$  

$x=\frac{(15\;mL)(230\;mL)}{180\;mL}\approx19.17\;mL$  
19.17 mL of medicine must be mixed in 230 mL of water. 
Conversion  Steps  Example 
Percent ⇒ Decimal  Move the decimal point two places to the left, then remove %.  31% = 31.% = 0.31 
Decimal ⇒ Percent  Move the decimal point two places to the right, then insert %.  0.317 = 0. 317 = 31.7 % 
Percent ⇒ Fraction  Remove %, divide by 100, then simplify.  15% = $\frac{15}{100}=\frac{3}{20}$ 
Fraction ⇒ Percent  Divide, move the decimal point two places to the right, then insert %.  $\frac{1}{4}$ = 1÷4 = 0.25 = 25 % 
Decimal ⇒ Fraction  Convert the decimal to a percent, then convert the percent to a fraction.  0.35 = 35 % = $\frac{35}{100}=\frac{7}{20}$ % = per one hundred 
$\frac{Part}{Whole}=\frac{Percent}{100}$  or  $\frac{"is"\;number}{"of"\;number}=\frac{\%}{100}$ 
Step  Example 
8 percent of what number is 4 ?  

⇑ Percent  ⇑ Whole (x)  ⇑ Part  

$\frac{4}{x}=\frac{8}{100}$  $\frac{Part}{Whole}=\frac{\%}{100}$  

$x=\frac{(4)(100)}{8}=50$  $x=50$ 
Example 1.4.4
Divide both sides by 90.
3) 12 is 8% of what number? 12 = 0.08 • x x = $\frac{12}{0.08}$ = 150Divide both sides by 0.08.
Application  Formula 
Percent increase  Percent increase = $\frac{New\;valueOriginal\;value}{Original\;value}$  $x=\frac{NO}{O}$ 
Percent decrease  Percent decrease = $\frac{Original\;valueNew\;value}{Original\;value}$  $x=\frac{ON}{O}$ 
Example 1.4.5
New value (N):  1650  This month. 
Original value (O):  1500  Last month. 
Percent increase:  $x=\frac{NO}{O}=\frac{16501500}{1500}=0.1=10\%}$  A 10% increase. 
Percent decrease:  $x=\frac{ON}{O}=\frac{3329}{33}\approx0.12=12\%$  A 12 % decrease. 
a. 4 nickels to 16 nickels.
b. 350 people for 1500 tickets.
c. 160 kilometres per 740 minutes.
2. A train traveled 459 km in 6 hours. What is the unit rate? 3. Write the following sentence as a proportion: 24 hours is to 1,940 kilometers as 12 hours is to 985 kilometers. 4. 4 liters of juice cost \$7.38, how much do 2 liters cost? 5. Sarah earns \$4,500 in 30 days. How much does she earn in 120 days? 6. A product increased production from 2,800 last year to 3,920 this year. Find the percent increase. 7. The table below gives survey data regarding Ryerson Occupational and Public Health students and summer jobs, but some data is missing:Year  2015  2016  2017 
Unemployed students  350  ?  396 
Total number of students  1250  1100  ? 
a. What % of students were unemployed in 2015?
b. If the % of unemployed students remained the same in 2016, find the number of unemployed students in 2016.
c. If the % of unemployed students increased by 5% in 2017, find the total number of students in 2017.
Answers
Exponential notation  Example 
Base Exponent  
a^{n}= a ∙ a ∙ a ∙ a … a  2^{4} = 2 ∙ 2 ∙ 2 ∙ 2 = 16 
Read “a to the nth” or “the nth power of a.”  Read “2 to the 4th.” 
Name  Rule  Example 
Product rule  $a^m\;a^n=a^{m+n}$  $2^3\;2^2=2^{3 + 2}=2^5=32$ 
Quotient rule  $\frac{a^m}{a^n}=a^{mn}$  $\frac{y^4}{y^2}=y^{42}=y^2$ 
Power rule  $(a^m)^n=a^{mn}$ $(a^m \cdot b^n)^p=a^{mp}\;b^{np}$ $(\frac{a^m}{b^n})^p=\frac{a^{mp}}{b^{np}}$  $(x^3)^2=x^{3\cdot2}=x^6$ $(t^3 \cdot s^4)^2=t^{3 \cdot 2}\;s^{4 \cdot 2}=t^6\;s^8$ $(\frac{q^2}{p^4})^3=\frac{q^{2\cdot3}}{p^{4\cdot3}}=\frac{q^6}{p^{12}}$ 
Negative exponent a^{n}  $a^{n}=\frac{1}{a^n}$  $4^{2}=\frac{1}{4^2}=\frac{1}{16}$ 
$\frac{1}{a^{n}}=a^n$  $\frac{1}{4^{2}}=4^2=16$  
Zero exponent a^{0}  $a^0=1$  $15^0=1$ 
One exponent a^{1}  $a^1=a$  $7^1=7$ , $1^{13}=1$ 
Fractional exponent  $a^\frac{m}{n}=\sqrt[n]{a^m}$  $15^\frac{2}{3}=\sqrt[3]{15^2}$ 
a^{m} a^{n} = a^{m}^{ + n} a^{n} or Base^{Exponent}
Example:  2^{3 }2^{2} = (2 · 2 · 2) (2 · 2) = 2^{5} = 3^{2}  
Or  2^{3 }2^{2 }= 2^{3 + 2} = 2^{5} = 32  A short cut, a^{m} a^{n} = a^{m}^{ + n} 
$\frac{a^m}{a^n}=a^{mn}$
Example:  $\frac{\bf2^4}{\bf2^2}=\frac{2\cdot2\cdot\bcancel{2\cdot2}}{\bcancel{2\cdot2}}=2^2=4$  
Or  $\frac{\bf2^4}{\bf2^2}=2^{42}=2^2=4$  A short cut, $\frac{a^m}{a^n}=a^{mn}$ 
(a^{m})^{n} = a^{mn}, (a^{m} · b^{n})^{p} = a^{mp} b^{np}, $(\frac{a^m}{b^n})^p=\frac{a^{mp}}{b^{np}}$
Example:  (4^{3})^{2} = (4^{3}) (4^{3}) = (4 · 4 · 4) (4 · 4 · 4) = 4^{6} = 4096  
Or  (4^{3})^{2} = 4^{3 ∙ 2} = 4^{6} = 4096  A short cut, (a^{m})^{n} = a^{mn} 
Example:  (2 · 3)^{2} = (2 · 3) (2 · 3) = 6 ∙ 6 = 36  
Or  (2 · 3)^{2 }= 2^{2} 3^{2} = 4 ∙ 9 = 36  A short cut , (a · b)^{n} = a^{n} b^{n} 
Example:  $(\frac{2^2}{3^4})^3=(\frac{2^2}{3^4})(\frac{2^2}{3^4})(\frac{2^2}{3^4})=\frac{4\cdot4\cdot4}{81\cdot81\cdot81}=\frac{64}{531441}$  
Or  $(\frac{2^2}{3^4})^3=\frac{2^{2\cdot3}}{3^{4\cdot3}}=\frac{2^6}{3^{12}}=\frac{64}{531441}$  A short cut, $(\frac{a^m}{b^n})^p=\frac{a^{mp}}{b^{np}}$ 
$a^{n}=\frac{1}{a^n}$, $\frac{1}{a^{n}}=a^n$ a^{−n} is the reciprocal of a^{n}.
Example:  $3^{4}=\frac{1}{3^4}=\frac{1}{81}$  $a^{n}=\frac{1}{a^n}$ 
Example:  $\frac{1}{3^{4}}=3^4=81$  $\frac{1}{a^{n}}=a^n$ 
$a^{\frac{m}{n}} = {\sqrt[n]{a}^{m}} = {\sqrt[n]{a^{m}}}$
Example:  $5^{\frac{3}{2}} = {\sqrt[2]{5}^{3}} = {\sqrt[2]{5^{3}}}$  $a^{\frac{m}{n}} = {\sqrt[n]{a^{m}}}$ 
Example 1.5.1
1) ${\bf (4)1}=4$  $a^1=a$ 
2) ${\bf (2345)^0}=1$  $a^0=1$ 
3) ${\bf x^2x^3}=x^{2+3}=x^5$  $a^m\;a^n=a^{m+n}$ 
4) ${\bf \frac{y^6}{y^4}}=y^{64}=y^2$  $\frac{a^m}{a^n}=a^{mn}$ 
5) ${\bf (x^4)^{3}}=x^{4(3)}=x^{12}=\frac{1}{x^{12}}$  $(a^m)^n=a^{mn}$ , $\frac{1}{a^{n}}=a^n$ 
6) ${\bf 7b^{1}}=7\cdot \frac{1}{b^1}=\frac{7}{b}$  $a^{n}=\frac{1}{a^n}$ , $a^1=a$ 
7) ${\bf (2t^3\cdot w^2)^4}=2^4 t^{3\cdot4}\cdot w^{2\cdot4}=16t^{12} w^8$  $(a^m \cdot b^n)^p=a^{mp}\;b^{np}$ 
8) ${\bf \frac{1}{3^{2}}}=3^2=9$  $\frac{1}{a^{n}}=a^n$ 
9) ${\bf \frac{7x^4y^{5}}{9^0\cdot x^2y^3}}=\frac{7x^{42}y^{53}}{1}=7x^2y^{8}=\frac{7x^2}{y^8}$  $a^0=1$ , $\frac{a^m}{a^n}=a^{mn}$ , $a^{n}=\frac{1}{a^n}$ 
10) ${\bf (\frac{e^{3}f^2}{g^{2}})^{2}}=\frac{e^{(3)(2)}f^{2(2)}}{g^{(2)(2)}}=\frac{e^6f^{4}}{g^4}=\frac{e^6}{g^4f^4}$  $(\frac{a^m}{b^n})^p=\frac{a^{mp}}{b^{np}}$ , $\frac{1}{a^{n}}=a^n$ 
Example 1.5.2
1) $\bf (3x^3y^2)^2 (2x^{3}y^{1})^3 (248z^{19})^0$  
$=3^2x^{3\cdot2}y^{2\cdot2} \cdot 2^3x^{3\cdot3} \cdot y^{1\cdot3}\cdot1$  Remove brackets.  $(\frac{a^m}{b^n})^p=\frac{a^{mp}}{b^{np}}$ , $a^0=1$ 
$=(3^2\cdot2^3)(x^6x^{9})(y^4y^{3})$  Regroup coefficients and variables.  
$=72x^3y^1$  Simplify.  $a^m\;a^n=a^{m+n}$ 
$=\frac{72y}{x^3}$  Make exponent positive.  $a^{n}=\frac{1}{a^n}$ , $a^1=a$ 
2) $\bf (\frac{(2x^4)(y^5)}{3x^3y^2})^2$  $(\frac{a^m}{b^n})^p=\frac{a^{mp}}{b^{np}}$  
$=\frac{(2x^4)^2(y^5)^2}{(3x^3y^2)^2}$  
$=\frac{2^2x^{4\cdot2}y^{5\cdot2}}{3^2x^{3\cdot2}y^{2\cdot2}}$  Remove brackets.  $(a \cdot b)^n=a^n\;b^n$ 
$=\frac{4}{9}\cdot \frac{x^8}{x^6}\cdot \frac{y^{10}}{y^4}$  Regroup coefficients and variables.  
$=\frac{4}{9}x^2y^6$  Simplify.  $\frac{a^m}{a^n}=a^{mn}$ 
Example 1.5.3
1) ${\bf (29a^{5}b^4c^{7})^0}=1$  $a^0=1$ 
2) ${\bf (\frac{a}{b})^{4}}=(\frac{2}{1})^{4}$  Substitute 2 for a and 1 for b, 
$=\frac{2^{4}}{1^{4}}=\frac{1^4}{2^4}=\frac{1}{16}$  $\frac{a^m}{a^n}=a^{mn}$ , $a^{n}=\frac{1}{a^n}$ , $\frac{1}{a^{n}}=a^n$ 
3) ${\bf (a+bc)^a}=[2+1(1)]^2=4^2=16$  Substitute 2 for a and 1 for b, and 1 for c. 
0.00000000000000000016 = 1.6 × 10^{19} C An electron.
Scientific notation  Example 
N × 10±n  1 ≤ N < 10  67504.3 = 6.75043 × 10^{4}  
n  integer  Standard form  Scientific notation 
Step  Example 

0.0079 37213000 

n = 3 n = 7 

0.0079 = 7.9 × 10^{3} 3 places to the right. 

37213000. = 3.7213 × 10^{7} 7 places to the left. 
Example 1.5.4
Example 1.5.5
a. 4x^{2} + 5y, for x = 1, y = 4
b. (2a)^{3} – 3b, for a = 5, b = 6
2. Simplify (do not leave negative exponents in the answer):a. (92)^{1}
b. y^{4} y^{3}
c. $\frac{x^9}{x^6}$
d. 13a^{1}
e. (3a^{2} · b^{3})^{4}
f. $\frac{5x^5y^{6}}{11^0\cdotx^3y^4}$
g. $(\frac{u^{2}v^3}{w^{4}})^3$
3. Write in scientific notation:a. 45,600,000
b. 0.00000523
4. Write in standard (or ordinary) form:a. 3.578 × 10^{3}
b. 4.3 × 10^{5}
Answers
Example 2.1.1
X  1  0  1 
Y  1  2  5 
Example 2.1.2
x  1  0  1 
y  6  4  2 
Example 2.1.3
Example 2.1.4
t  0  1  2 
x  3  5  7 
y  1  2  3 
Example 2.1.5
Answers
Example 3.1.1
Example 3.1.2
Example 3.1.3
“height is a function of age”  if we name the function f we write 
“h is f of a”  or more simply 
h = f(a)  we could instead name the function h and write 
h(a)  which is read “h of a” 
Example 3.1.4
Example 3.1.5
(input) Month number, m  1  2  3  4  5  6  7  8  9  10  11  12 
(output) Days in month, D  31  28  31  30  31  30  31  31  30  31  30  31 
n  1  2  3  4  5 
Q  8  6  7  6  8 
(input) a, age in years  5  5  6  7  8  9  10 
(output) h, height inches  40  42  44  47  50  52  54 
Example 3.1.6
Input  Output  Input  Output  Input  Output 
2  1  3  5  1  0 
5  3  0  1  5  2 
8  6  4  5  5  4 
Example 3.1.7
n  1  2  3  4  5 
Q  8  6  7  6  8 
Example 3.1.8
Example 3.1.9
Example 3.1.10
Example 3.1.11
= 1 + 2
= 1 which was the desired result.
Example 3.1.12
= 16 + 8
= 24
b) h(p) = 3 Substitute the original function h(p) = p^{2} + 2p p^{2} + 2p = 3 This is quadratic, so we can rearrange the equation to get it = 0 p^{2} + 2p – 3 = 0 subtract 3 from each side p^{2} + 2p – 3 = 0 this is factorable, so we factor it (p + 3)(p – 1) = 0 By the zero factor theorem since (p + 3)(p – 1) = 0, either (p + 3) = 0 or (p – 1) = 0 (or both of them equal 0) and so we solve both equations for p, finding p = 3 from the first equation and p = 1 from the second equation. This gives us the solution: h(p) = 3 when p = 1 or p = 3.Answers
Example 3.4.1
Example 3.4.2
x  g(x) 
5  1.5 
4  1 
3  2.5 
2  3 
1  2.5 
0  1 
1  1.5 
a > 0 
a < 0 
Example 3.4.3
$g(x)=\frac{1}{2}(x+2)^23$
$g(x)=\frac{1}{2}(x+2)(x+2)3$
$g(x)=\frac{1}{2}(x^2+4x+4)3$
$g(x)=\frac{1}{2}x^2+2x+23$
$g(x)=\frac{1}{2}x^2+2x1$
$h=\frac{b}{2a}$, $k=f(h)=f(\frac{b}{2a})$
Example 3.4.4
$f(x)=2(x\frac{3}{2})^2+\frac{5}{2}$
Example 3.4.5
zero horizontal intercepts  one horizontal intercept  two horizontal intercepts 
$x=\frac{b\pm\sqrt{b^24ac}}{2a}}$
Example 3.4.6
Example 3.4.7
a. $g(x)=8x+x^2+7$
b. $g(x)=3(3x)^2+2$
4. A ball is thrown from the top of a 35 m tall building at a speed of 24.5 m per second. The ball’s height above ground can be modelled by the equation: $h(t)=4.9t^2+24.5t+35$.a. What is the maximum height of the ball?
b. When does it hit the ground?
Answers
$h(x)=\frac{7}{16}(x+4)^2+7&$ .
To make the shot, h(7.5) would need to be about 4.
h(7.5) ≈ 1.64; he doesn't make it.
2. $g(x)=x^26x+13$ in standard form; $g(x)=(x3)^2+4$ in vertex form. 3. a. Vertex is a minimum value (opens upwards). b. Vertex is a maximum value (opens downwards). 4. a. $65.625$ m b. $\approx 6.16$ secYear  Stores, Company A  Stores, Company B 
0  100  Starting with 100 each  100 
1  100 + 50 = 150  They both grow by 50 stores in the first year.  100 + 50% of 100 100 + 0.50(100) = 150 
2  150 + 50 = 200  Store A grows by 50, Store B grows by 75  150 + 50% of 150 150 + 0.50(150) = 225 
3  200 + 50 = 250  Store A grows by 50, Store B grows by 112.5  225 + 50% of 225 225 + 0.50(225) = 337.5 
Year  Company A  Company B 
2  200  225  
4  300  506  
6  400  1139  
8  500  2563  
10  600  5767 
Example 4.1.1
Example 4.1.2
Example 4.1.3
Example 4.1.4
Example 4.1.5
x  3  2  1  0  1  2  3 
f(x)  $\frac{1}{8}$  $\frac{1}{4}$  $\frac{1}{2}$  1  2  4  8 
x  3  2  1  0  1  2  3 
g(x)  8  4  2  1  $\frac{1}{2}$  $\frac{1}{4}$  $\frac{1}{8}$ 
$f(x)=2^{x}=(2^{1})^x=(\frac{1}{2})^x=g(x)$
Looking at the graphs also confirms this relationship: Consider a function for the form f(x) = ab^{x} . Since a, which we called the initial value in the last section, is the function value at an input of zero, a will give us the vertical intercept of the graph. From the graphs above, we can see that an exponential graph will have a horizontal asymptote on one side of the graph, and can either increase or decrease, depending upon the growth factor. A horizontal asymptote is a horizontal line y = b where the graph approaches the line as the inputs get large. The horizontal asymptote will help us determine the long run behaviour and is easy to determine from the graph. The graph will grow when the growth rate is positive, which will make the growth factor b larger than one. When it’s negative, the growth factor will be less than one.Example 4.1.6
Example 4.1.7
$f(x)=2(1.3)^x$ $g(x)=2(1.8)^x$ $h(x)=4(1.3)^x$ $k(x)=4(0.7)^x$ 
Example 4.1.8
Example 4.1.9
Frequency  Value after 1 year 
Annually  \$1100 
Semiannually  \$1102.50 
Quarterly  \$1103.81 
Monthly  \$1104.71 
Daily  \$1105.16 
Frequency  Value 
Annually  \$2 
Semiannually  \$2.25 
Quarterly  \$2.441406 
Monthly  \$2.613035 
Daily  \$2.714567 
Hourly  \$2.718127 
Once per minute  \$2.718279 
Once per second  \$2.718282 
Example 4.1.10
$A(t)=1000(1.05)^t$ $B(t)=900(1.075)^t$
3. A population of 1000 is decreasing 3% each year. Find the population in 30 years. 4. Recalculate Example 4.1.8 with monthly compounding. 5. The population of a nest of ants is 225, with a continuous growth rate of 18% per week. What is the population size of the nest after 8 weeks? 6. You are told that your 42 yearold hockey card is worth \$1200. If this card appreciated at 19% per year compounded semiannually, what was the initial value of the card?Answers
C ={Ken, Bob, Tran, Shanti, Eric}
A set that has no members is called an empty set. The empty set is denoted by the symbol Ø. Two sets are equal if they have the same elements. A set A is a subset of a set B if every member of A is also a member of B. Suppose C = {Al, Bob, Chris, David, Ed} and A = {Bob, David}. Then A is a subset of C, written as $A\subseteq C$. Every set is a subset of itself, and the empty set is a subset of every set.Example 5.1.1
Example 5.1.2
Example 5.1.3
Example 5.1.4
Example 5.1.5
Example 5.1.6
\begin{tikzpicture}
[thick]
\draw (0,0) circle (1) node[below,shift={(1,1.1)}] {$\bm{A}$};
\draw (1.2,0) circle (1) node[below,shift={(1,1.1)}] {$\bm{S}$};\node at (.6,0) {12};\end{tikzpicture}
(a) 
\begin{tikzpicture}[thick]
\draw (0,0) circle (1) node[below,shift={(1,1.1)}] {$\bm{A}$};
\draw (1.2,0) circle (1) node[below,shift={(1,1.1)}] {$\bm{S}$};\node at (.6,0) {12};\node at (.3,0) {18};
\end{tikzpicture}
(b) 
\begin{tikzpicture}[thick]
\draw (0,0) circle (1) node[below,shift={(1,1.1)}] {$\bm{A}$};
\draw (1.2,0) circle (1) node[below,shift={(1,1.1)}] {$\bm{S}$};\node at (.6,0) {12};
\node at (1.5,0) {8};
\node at (.3,0) {18};
\end{tikzpicture}
(c) 
Example 5.1.7
\begin{tikzpicture}[thick]
\draw (2.7,1.5) rectangle (1.5,1.5) node[below,shift={(3.9,0)}] {$\bm{U}$};
\draw (0,0) circle (1) node[below,shift={(1.2,0.8)}] {$\bm{D}$};
\draw (1.2,0) circle (1) node[below,shift={(1.2,0.8)}] {$\bm{K}$};\node at (.6,0) {6};
\node at (1.5,0) {9};
\node at (.3,0) {54};
\node at (2.3,0.8) {x};
\end{tikzpicture}
(a) 
\begin{tikzpicture}[thick]
\draw (2.7,1.5) rectangle (1.5,1.5) node[below,shift={(3.8,0)}] {$\bm{U}$};
\draw (0,0) circle (1) node[below,shift={(1.2,0.8)}] {$\bm{D}$};
\draw (1.2,0) circle (1) node[below,shift={(1.2,0.8)}] {$\bm{K}$};\node at (.6,0) {6};
\node at (1.5,0) {9};
\node at (.3,0) {54};
\node at (2.3,0.8) {31};
\end{tikzpicture}
(b) 
We fill the three regions associated with the sets D and K in the same manner as before. Since 100 people participated in the survey, the rectangle representing the universal set U must contain 100 objects. Let x represent those people in the universal set that are neither in the set D nor in K. This means 54 + 6 + 9 + x = 100, or x = 31.
Therefore, there are 31 people in the survey who have visited neither place.
Example 5.1.8
\begin{tikzpicture}[thick]
\draw (2.7,2.6) rectangle (1.5,1.5) node[below,shift={(3.9,0)}] {$\bm{U}$};
\draw (0,0) circle (1) node[below,shift={(1.2,0.8)}] {$\bm{J}$};
\draw (1.2,0) circle (1) node[below,shift={(1.2,0.8)}] {$\bm{S}$};\draw (0.6,1.1) circle (1) node[below,shift={(1.1,0.4)}] {$\bm{C}$};\node at (.6,0.4) {3};\end{tikzpicture}
(a) 
\begin{tikzpicture}[thick]
\draw (2.7,2.6) rectangle (1.5,1.5) node[below,shift={(3.9,0)}] {$\bm{U}$};
\draw (0,0) circle (1) node[below,shift={(1.2,0.8)}] {$\bm{J}$};
\draw (1.2,0) circle (1) node[below,shift={(1.2,0.8)}] {$\bm{S}$};\draw (0.6,1.1) circle (1) node[below,shift={(1.1,0.4)}] {$\bm{C}$};\node at (.6,0.4) {3};
\node at (1.1,0.7) {4};
\node at (0.1,0.7) {6};
\node at (0.6,0.2) {11};\end{tikzpicture}
(b) 
\begin{tikzpicture}[thick]
\draw (2.7,2.6) rectangle (1.5,1.5) node[below,shift={(3.9,0)}] {$\bm{U}$};
\draw (0,0) circle (1) node[below,shift={(1.2,0.8)}] {$\bm{J}$};
\draw (1.2,0) circle (1) node[below,shift={(1.2,0.8)}] {$\bm{S}$};\draw (0.6,1.1) circle (1) node[below,shift={(1.1,0.4)}] {$\bm{C}$};\node at (.6,0.4) {3};\node at (1.1,0.7) {4};
\node at (0.1,0.7) {6};
\node at (0.6,0.2) {11};
\node at (1.5,0) {12};
\node at (.3,0) {30};
\node at (2.3,2.2) {12};\node at (0.6,1.5) {22};\end{tikzpicture}
(c) 
We place a 3 in the innermost region of Figure (a) because it represents the number of people who participate in all three activities. Next we compute x, y and z.
a. The number of students subscribing to Amazon Prime but not the other two streaming services.
b. The number of students subscribing to Netflix or Amazon Prime but not Crave.
c. The number of students not subscribing to any of these services.
Answers
HH HT TH TT
The possibility HT, for example, indicates a head on the penny and a tail on the nickel, while TH represents a tail on the penny and a head on the nickel. It is for this reason, we emphasize the need for understanding sample spaces. An act of flipping coins, rolling dice, drawing cards, or surveying people are referred to as an experiment.Example 6.1.1
Example 6.1.2
Example 6.1.3
Green  
Red  1  2  3  4  5  6 
1  (1, 1)  (1, 2)  (1, 3)  (1, 4)  (1, 5)  (1, 6) 
2  (2, 1)  (2, 2)  (2, 3)  (2, 4)  (2, 5)  (2, 6) 
3  (3, 1)  (3, 2)  (3, 3)  (3, 4)  (3, 5)  (3, 6) 
4  (4, 1)  (4, 2)  (4, 3)  (4, 4)  (4, 5)  (4, 6) 
5  (5, 1)  (5, 2)  (5, 3)  (5, 4)  (5, 5)  (5, 6) 
6  (6, 1)  (6, 2)  (6, 3)  (6, 4)  (6, 5)  (6, 6) 
Example 6.1.4
Example 6.1.5
Example 6.1.6
Example 6.1.7
Example 6.1.8
a. P (a club)
b. P (a jack or spade)
3. A jar contains 6 red, 7 white, and 7 blue marbles. If a marble is chosen at random, find the following probabilities:a. P (red)
b. P (red or blue)
4. Two dice are rolled. Find the following probabilities:a. P (the sum of the dice is 5)
b. P (the sum of the dice is 3 or 6)
Answers
Example 7.1.1
Outcome  Four Hits  Three hits  Two Hits  One hits  No Hits 
Probability  (.3)^{4}  4(.3)^{3}(.7)  6(.3)^{2}(.7)^{2}  4(.3)(.7)^{3}  (.7)^{4} 
P(n, k; p) = nCkp^{k}q^{n  k}
where p denotes the probability of success and q = (1 − p) the probability of failure.Example 7.1.2
Example 7.1.3
Example 7.1.4
Example 7.1.5
Example 7.1.6
Answers
Example 7.4.1
Example 7.4.2
Monday  Tuesday  Wednesday 
The probability that Professor Symons walked on Wednesday given that he walked on Monday can be found from the tree diagram, as listed below.
P(Walked Wednesday  Walked Monday) = P(WWW) + P(WBW) = 1/4 + 1/8 = 3/8.
P(Bicycled Wednesday  Walked Monday) = P(WWB) + P(WBB) = 1/4 + 3/8 = 5/8.
P(Walked Wednesday  Bicycled Monday) = P(BWW) + P(BBW) = 1/8 + 3/16 = 3/5/16.
P(Bicycled Wednesday  Bicycleed Monday) = P(BWB) + P(BBB) = 1/8 + 9/16 = 11/16.
We represent the results in the following matrix:
Alternately, this result can be obtained by squaring the original transition matrix.
We list both the original transition matrix T and T^{2} as follows:
$T= \left[ \begin{array}{cc} 1/2 & 1/2 \\ 1/4 & 3/4 \end{array} \right]$
$T^2\quad\quad \left[ \begin{array}{cc} 1/2 & 1/2 \\ 1/4 & 3/4 \end{array} \right] \left[ \begin{array}{cc} 1/2 & 1/2 \\ 1/4 & 3/4 \end{array} \right]$
$\left[ \begin{array}{cc} (1/2)(1/2) + (1/2)(1/4) & (1/2)(1/2) + (1/2)(3/4) \\ (1/4)(1/2) + (3/4)(1/4) & (1/4)(1/2) + (3/4)(3/4) \end{array} \right]$
$\left[ \begin{array}{cc} 1/4 + 1/8 & 1/4 + 3/8 \\ 1/8 + 3/16 & 1/8 + 9/16 \end{array} \right]$
$\left[ \begin{array}{cc} 3/8 & 5/8 \\ 5/16 & 11/16 \end{array} \right]$
The reader should compare this result with the probabilities obtained from the tree diagram. Consider the following case, for example:
P(Walked Wednesday  Bicycled Monday) = P(BWW) + P(BBW) = 1/8 + 3/16 = 5/16.
It makes sense because to find the probability that Professor Symons will walk on Wednesday given that he bicycled on Monday, we sum the probabilities of all paths that begin with B and end in W. There are two such paths, and they are BWW and BBW.
Example 7.4.3
ET = E
or,
$\left[\begin{array}{cc} 1/3 & 2/3\end{array}\right] \left[ \begin{array}{cc} 1/3 & 2/3 \\ 1/3 & 2/3 \end{array}\right]=\left[\begin{array}{cc} 1/3 & 2/3\end{array}\right]$
Example 7.4.4
If the initial market share for Mama Bell is 20% and for Papa Bell 80%, we'd like to know the long term market share for each company.
Let matrix T denote the transition matrix for this Markov chain, and M denote the matrix that represents the initial market share. Then T and M are as follows:
$T=\left[\begin{array}{cc} 0.60 & 0.40 \\0.30 & 0.70\end{array}\right]$ and $M=\left[\begin{array}{cc} 0.20 & 0.80\end{array}\right]$Since each month the town's people switch according to the transition matrix T, after one month the distribution for each company is as follows:
$\left[\begin{array}{cc} 0.20 & 0.80\end{array}\right] \left[\begin{array}{cc} 0.60 & 0.40 \\0.30 & 0.70\end{array}\right]=\left[\begin{array}{cc} (0.20)(0.60) + (0.80)(0.30) & (0.20)(0.40) + (0.80)(0.70)\end{array}\right]=\left[\begin{array}{cc} 0.36 & 0.64\end{array}\right]$
After two months, the market share for each company is:
$\left[\begin{array}{cc} 0.36 & 0.64\end{array}\right] \left[\begin{array}{cc} 0.60 & 0.40 \\0.30 & 0.70\end{array}\right]=\left[\begin{array}{cc} 0.408 & 0.592\end{array}\right]$
After three months the distribution is:
$\left[\begin{array}{cc} 0.408 & 0.592\end{array}\right] \left[\begin{array}{cc} 0.60 & 0.40 \\0.30 & 0.70\end{array}\right]=\left[\begin{array}{cc} 0.4224 & 0.5776\end{array}\right]$
After four months the market share is:
$\left[\begin{array}{cc} 0.4224 & 0.5776\end{array}\right] \left[\begin{array}{cc} 0.60 & 0.40 \\0.30 & 0.70\end{array}\right]=\left[\begin{array}{cc} 0.42672 & 0.57328\end{array}\right]$
After 30 months the market share is $\left[\begin{array}{cc} 3/7 & 4/7\end{array}\right]$ .
The market share after 30 months has stabilized to $\left[\begin{array}{cc} 3/7 & 4/7\end{array}\right]$ .
This means that:
$\left[\begin{array}{cc} 3/7 & 4/7\end{array}\right] \left[\begin{array}{cc} .60 & .40 \\.30 & .70\end{array}\right]=\left[\begin{array}{cc} 3/7 & 4/7\end{array}\right]$
Once the market share reaches an equilibrium state, it stays the same, that is, ET = E.
Example 7.4.5
a. What is the percentage of people in the community that reported texting and driving after the campaign?
b. If the campaign were to be repeated multiple times, what is the longrange trend in terms of the lowest rate that texting and driving can be reduced to in this community?
Answers
$\frac{rise}{run},\frac{vertical\;change}{horizontal\;change},\frac{\Delta y}{\Delta x}etc.$
We give a precise definition. Definition: If ( x_{1}, y_{1}) and ( x_{2}, y_{2}) are two different points on a line, then the slope of the line is $Slope=m=\frac{y_2y_1}{x_2x_1}$Example 2.2.1
$m=\frac{3(1)}{24}=\frac{4}{6}=\frac{2}{3}$
The student should further observe that if a line rises when going from left to right, then it has a positive slope; and if it falls going from left to right, it has a negative slope.Example 2.2.2
Note: The slope of a vertical line is undefined.
Example 2.2.3
Example 2.2.4
Example 2.2.5
The line  Slope  yintercept 
y = 3x + 2  3  2 
y = 2x + 5  2  5 
y = 3/2x – 4  3/2  4 
Example 2.2.6
a. (2, 3) and (5, 9)
b. (6, 5) and (4, 1)
2. Determine the slope of the line from the given equation of the line:a. $y = 2x + 1$
b. $3x  4y = 12$
Answers
Example 2.3.1
Example 2.3.2
Example 2.3.3
Example 2.3.4
Example 2.3.5
Example 2.3.6
$m=\frac{0(2)}{41}=\frac{2}{3}$
The pointslope form is:$y(2)=2/3(x1)$
Multiplying both sides by 3 gives us:$3(y+2)=2(x1)$
$3y+6=2x2$
$2x+3y=8$
$2x3y=8$
Example 2.3.7
$3y = 2x + 9$
$2x + 3y = 9$
Example 2.3.8
Example 2.3.9
Example 2.3.10
a. Passes through (3, 5) and (2, 1).
b. Passes through (5, 2) and m = $2/5$.
c. Passes through (2, 5) and its xintercept is 4.
2. Write an equation of the line that passes through (4, 2) and m = $3/4$. Write the equation in the form Ax+By = C.Answers
Example 2.4.1
Example 2.4.2
Example 2.4.3
$m=\frac{1000750}{5025}=10$
Therefore, the partial equation is y = 10x + b. By substituting one of the points in the equation, we get b = 500. Therefore, the cost equation is y = 10x + 500. Now to find the cost of 100 items, we substitute x = 100 in the equation y = 10x + 500. So the cost = y = 10(100) + 500 = 1500.Example 2.4.4
Centigrade  Fahrenheit 
0  32 
100  212 
slope $m=\frac{21232}{1000}=\frac{9}{5}$
The equation is $F=\frac{9}{5}C+b$ Substituting the point (0, 32), we get$F=\frac{9}{5}C+32$
Now to convert 30 degrees Celsius into Fahrenheit, we substitute C = 30 in the equation$F=\frac{9}{5}C+32$
$F=\frac{9}{5}(30)+32=86$
Example 2.4.5
$m=\frac{2618}{160}=\frac{1}{2}$
The equation is $y=\frac{1}{2}x+b$. Substituting the point (0, 18), we get:$y=\frac{1}{2}x+18$
Now to find the population in the year 2010, we let x = 40 in the equation:$y=\frac{1}{2}x+18$
$y=\frac{1}{2}(40)+18=38$
So the population of Canada in the year 2010 is estimated as 38 million.Year  Population 
0 (1970)  18 million 
16 (1986)  26 million 
Example 2.4.6
$3x1=x+7$
$4x = 8$
$x = 2$
By substituting x = 2 in any of the two equations, we obtain y = 5. Hence, the solution (2, 5). One common algebraic method used in solving systems of equations is called the elimination method. The object of this method is to eliminate one of the two variables by adding the left and right sides of the equations together. Once one variable is eliminated, we get an equation that has only one variable for which it can be solved. Finally, by substituting the value of the variable that has been found in one of the original equations, we get the value of the other variable. The method is demonstrated in the example below.Example 2.4.7
Example 2.4.8
$2x4y=6$
$2x+3y=4$
$y=2$
$y = 2$
Substituting y = 2 in x + 2y = 3, we get$x+2(2) = 3$
$x =1$
Therefore, the solution is (1, 2).Example 2.4.9
Example 2.4.10
Example 2.4.11
Answers
Circumference, c  1.7  2.5  5.5  8.2  13.7 
Height, h  24.5  31  45.2  54.6  92.1 
Example 3.2.1
Example 3.2.2
Letters  
Weight not Over  Price 
1 ounce  \$0.44 
2 ounces  \$0.61 
3 ounces  \$0.78 
3.5 ounces  \$0.95 
Inequality  Set Builder Notation  Interval notation 
5 < h ≤ 10  {h  5 < h ≤ 10}  (5, 10] 
5 ≤ h < 10  {h  5 ≤ h < 10}  [5, 10) 
5 < h ≤ 10  {h  5 < h < 10}  (5, 10) 
h < 10  {h  h < 10}  (∞, 10) 
h ≥ 10  {h  h ≥ 10}  [10, ∞) 
all real numbers  {h  h ∈ ℜ}  (∞, ∞) 
Example 3.2.3
Example 3.2.4
Example 3.2.5
Example 3.2.6
Example 3.2.7
Example 3.2.8
Answers
t  2  3  4  5  6  7  8  9 
C(t)  1.47  1.69  1.94  2.30  2.51  2.64  3.01  2.14 
Example 3.3.1
Example 3.3.2
Example 3.3.3
t (hours)  0  1  2  3  4  5  6  7 
D(t) (miles)  10  55  90  153  214  240  292  300 
Example 3.3.4
Example 3.3.5
Average rate of change = $\frac{F(6)F(2)}{62}$  Evaluating the function 
$\frac{F(6)F(2)}{62}=$  
$\frac{\frac{2}{6^2}\frac{2}{2^2}}{62}$  Simplifying 
$\frac{\frac{2}{36}\frac{2}{4}}{4}$  Combining the numerator terms 
$\frac{\frac{16}{36}}{4}$  Simplifying further 
$\frac{1}{9}$ Newtons per centimeter 
Example 3.3.6
Example 3.3.7
t  2  3  4  5  6  7  8  9 
C(t)  1.47  1.69  1.94  2.30  2.51  2.64  3.01  2.14 
Answers
$=\frac{a^3+3a^2h+3ah^2+h^3+2a^32}{h}=\frac{3a^2h+3ah^2+h^3}{h}$
$=\frac{h(3a^2+3ah+h^2)}{h}=3a^2+3ah+h^2$
4. Based on the graph, the local maximum appears to occur at (1, 28), and the local minimum occurs at (5,80). The function is increasing on (∞, 1)∪(5, ∞) and decreasing on (1, 5).Example 4.2.1
Example 4.2.2
Example 4.2.3
Example 4.2.4
Values of the Common Log  
Number  Number as exponential  Log(number) 
1000  10^{3}  3 
100  10^{2}  2 
10  10^{1}  1 
1  10^{0}  0 
0.1  10^{1}  1 
0.01  10^{2}  2 
0.001  10^{3}  3 
Example 4.2.5
x  3  2  1  0  1  2  3 
f(x)  $\frac{1}{8}$  $\frac{1}{4}$  $\frac{1}{2}$  1  2  4  8 
x  $\frac{1}{8}$  $\frac{1}{4}$  $\frac{1}{2}$  1  2  4  8 
f(x)  3  2  1  0  1  2  3 
Example 4.2.6
Example 4.2.7
Example 4.2.8
Example 4.2.9
Example 4.2.10
Example 4.2.11
$2 = 1.14(1.0134)^t$  Divide by 1.14 to isolate the exponential expression 
$\frac{2}{1.14}=1.0134^t$  Take the logarithm of both sides of the equation 
$\ln(\frac{2}{1.14})=\ln(1.0134^t)$  Apply the exponent property on the right side 
$\ln(\frac{2}{1.14})=t\ln(1.0134)$  Divide both sides by ln(1.0134) 
Example 4.2.12
log_{3}(5) + log_{3}(8) = log_{3}(5 · 8) = log_{3}(40)
This reduces our original expression to log_{3}(40) – log_{3}(2) Then using the difference of logs property,log_{3}(40) – log_{3}(2) = log_{3}($\frac{40}{2}$) = log_{3}(20)
Example 4.2.13
2log(5) = log(5^{2}) = log(25)
With the expression reduced to a sum of two logs, log(25) + log(4) , we can utilize the sum of logs property:log(25) + log(4) = log(4 · 25) = log(100)
Since 100 = 10^{2}, we can evaluate this log without a calculator:log(100) = log(10^{2}) = 2
Example 4.2.14
$\ln(\frac{x^4y}{7})=\ln(x^4y)\ln(7)$
Then seeing the product in the first term, we use the sum property:$\ln(x^4y)\ln(7)=\ln(x^4)+\ln(y)\ln(7)$
Finally, we could use the exponent property on the first term:$\ln(x^4)+\ln(y)\ln(7)=4\ln(x)+\ln(y)\ln(7)$
Example 4.2.15
Answers
Example 4.3.1a
$Kenya(t)=38.8(1+0.0264)^t$
$Sudan(t)=41.3(1+0.0224)^t$
To find when the populations will be equal, we can set the equations equal:$38.8(1.0264)^t=41.3(1.0224)^t$
For our first approach, we take the log of both sides of the equation:$\log(38.8(1.0264)^t)=\log(41.3(1.0224)^t)$
Utilizing the sum property of logs, we can rewrite each side:$\log(38.8)+\log(1.0264^t)=\log(41.3)+\log(1.0224^t)$
Then utilizing the exponent property, we can pull the variables out of the exponent:$\log(38.8)+t\log(1.0264)=\log(41.3)+t\log(1.0224)$
Moving all the terms involving t to one side of the equation and the rest of the terms to the other side:$t\log(1.0264)t\log(1.0224)=\log(41.3)\log(38.8)$
Factoring out the t on the left:$t(\log(1.0264)\log(1.0224))=\log(41.3)\log(38.8)$
Dividing to solve for t:$t=\frac{\log(41.3)\log(38.8)}{\log(1.0264)\log(1.0224)}\approx15.991$ years until the populations will be equal.
Example 4.3.1b
38.8(1.0264)^{t} = 41.3(1.0224)^{t}
Divide to move the exponential terms to one side of the equation and the constants to the other side:$\frac{1.0264^t}{1.0224^t}=\frac{41.3}{38.8}$
Using exponent rules to group on the left:$(\frac{1.0264}{1.0224})^t = \frac{41.3}{38.8}$
Taking the log of both sides:$\log((\frac{1.0264}{1.0224})^t)=\log(\frac{41.3}{38.8})$
Utilizing the exponent property on the left:$t\log(\frac{1.0264}{1.0224})=\log(\frac{41.3}{38.8})$
Dividing gives:$t=\frac{\log(\frac{41.3}{38.8})}{\log(\frac{1.0264}{1.0224})}\approx15.991$ years
While the answer does not immediately appear identical to that produced using the previous method, note that by using the difference property of logs, the answer could be rewritten:$t=\frac{\log(\frac{41.3}{38.8})}{\log(\frac{1.0264}{1.0224})}=\frac{\log(41.3)\log(38.8)}{\log(1.0264)\log(1.0224)}$
Example 4.3.2
$\frac{1}{2}a=a(0.87)^d$  
$\frac{1}{2}=0.87^d$  Dividing by a 
$\log(\frac{1}{2})=\log(0.87^d)$  Taking the log of both sides 
$\log(\frac{1}{2})=d\log(0.87)$  Use the exponent property of logs 
$d=\frac{\log(\frac{1}{2})}{\log(0.87)}\approx4.977$ days  Divide to solve for d 
Example 4.3.3
Example 4.3.4
$\frac{1}{2}a=ae^{r5730}$  
$\frac{1}{2}=e^{r5730}$  Dividing by a 
$\ln(\frac{1}{2})=\ln(e^{r5730})$  Taking the natural log of both sides 
$\ln(\frac{1}{2})=5730r$  Use the inverse property of logs on the right side 
$r=\frac{\ln(\frac{1}{2})}{5730}\approx0.000121$  Dividing by 5730 
Example 4.3.5
Example 4.3.6
Answers
Example 7.2.1
\begin{tikzpicture}[thick]
\tikzstyle{black} = [circle, minimum width=8pt,fill,inner sep=0pt]
\tikzstyle{white} = [thin, circle, draw, minimum width=8pt,fill=white, inner sep=0pt]
\draw (0,0) circle (1);
\draw (2.5,0) circle (1);
\node at (0,1.3) {Jar I};
\node at (2.5,1.3) {Jar II};
\node at (0,0.7) [black]{};
\node at (0.4,0.6) [white] {};\node at (0.4,0.6) [white] {};\node at (0.7,0.3) [white] {};\node at (0.7,0.3) [white] {};\node at (2.5,0.7) [black] {};
\node at (2.9,0.6) [black] {};\node at (2.1,0.6) [black] {};\node at (3.2,0.3) [black] {};\node at (1.8,0.3) [white] {};\node at (2.7,0.3) [white] {};\node at (2.3,0.3) [white] {};\node at (2.9,0) [white] {};\node at (2.5,0) [white] {};\node at (2.1,0) [white] {};\draw (0,0.7) ellipse (6mm and 3mm);
\draw (2.5,0.7) ellipse (6mm and 3mm);
\end{tikzpicture}
(a) 
\begin{tikzpicture}[grow=right, sloped]
\tikzstyle{level 1}=[level distance=3cm, sibling distance=3.5cm]
\tikzstyle{level 2}=[level distance=3cm, sibling distance=2cm]
\node {}
child {
node {$JII$}
child {
node[label=right:
{$White\: (1/2)(6/10)=3/10$}] {}
edge from parent
node[below] {$6/10$}
}
child {
node[label=right:
{$Black\: (1/2)(4/10)=2/10$}] {}
edge from parent
node[above] {$4/10$}
}
edge from parent
node[below] {$1/2$}
}
child {
node {$JI$}
child {
node[label=right:
{$White\: (1/2)(4/5)=4/10 $}] {}
edge from parent
node[below] {$4/5$}
}
child {
node[label=right:
{$Black\: (1/2)(1/5)=1/10$}] {}
edge from parent
node[above] {$1/5$}
}
edge from parent
node[above] {$1/2$}
};
\end{tikzpicture}
(b) 
$P(A_i\,\, E) = \frac{P(A_i)\, P(E\,\, A_i)}{P(A_1)\, P(E\,\, A_1)+P(A_2)\, P(E\,\, A_2)+\,\cdots\,+P(A_n)\, P(E\,\, A_n)}$
Example 7.2.2
The probability P(A  R), for example, is a fraction whose denominator is the sum of all probabilities of all branches of the tree that result in an appliance that needs repair before the warranty expires, and the numerator is the branch that is associated with Manufacturer A. P(B  R) and P(C  R) are found in the same way. We list both as follows:
Example 7.2.3
Store Number  Number of Employees  Percent of Women Employees 
1  300  .40 
2  150  .65 
3  200  .60 
4  250  .50 
5  100  .70 
Total = 1000 
Example 7.2.4
Positive test  Negative test  Total  
Have disease  180  20  200 
Do not have disease  98  9,702  9,800 
Total  278  9,822  10,000 
a. P (marble is red)
b. P (The marble came from Jar II given that a white marble is drawn)
c. P (Red marble  Jar I)
2. A certain disease has an incidence rate of 0.5%. If there are no false negatives and if the false positive rate is 3%, compute the probability that a person who tests positive actually has the disease. Assume the test is applied to 100,000 people. 3. A computer company buys its chips from three different manufacturers. Manufacturer I provides 60% of the chips, of which 5% are known to be defective; Manufacturer II supplies 30% of the chips, of which 4% are defective; while the rest are supplied by Manufacturer III, of which 3% are defective. If a chip is chosen at random, find the following probabilities:a. P (The chip is defective)
b. P (The chip came from Manufacturer II  it is defective)
c. P (The chip is defective  it came from manufacturer III)
Answers
1. Suppose you and your friend play a game that consists of rolling a die. Your friend offers you the following deal: If the die shows any number from 1 to 5, he will pay you the face value of the die in dollars, that is, if the die shows a 4, he will pay you \$4. But if the die shows a 6, you will have to pay him \$18.
Before you play the game you decide to find the expected value. You analyze as follows.
Since a die will show a number from 1 to 6, with an equal probability of 1/6, your chance of winning \$1 is 1/6, winning \$2 is 1/6, and so on up to the face value of 5. But if the die shows a 6, you will lose \$18. You write the expected value.
E = \$1(1/6) + \$2(1/6) + \$3(1/6) + \$4(1/6) + \$5(1/6) − \$18(1/6) = −\$0.50
This means that every time you play this game, you can expect to lose 50 cents. In other words, if you play this game 100 times, theoretically you will lose \$50. Obviously, it is not to your interest to play.
2. Suppose of the 10 quizzes you took in a course, on eight quizzes you scored 80, and on two you scored 90. You wish to find the average of the 10 quizzes. The average is:
$A=\frac{(80)(8)+(90)(2)}{10}=(80)\frac{8}{10}+(90)\frac{2}{10}=82$
It should be observed that it will be incorrect to take the average of 80 and 90 because you scored 80 on eight quizzes, and 90 on only two of them. Therefore, you take a "weighted average" of 80 and 90. That is, the average of 8 parts of 80 and 2 parts of 90, which is 82.
In the first situation, to find the expected value, we multiplied each payoff by the probability of its occurrence, and then added up the amounts calculated for all possible cases. In the second example, if we consider our test score a payoff, we did the same. This leads us to the following definition.Payoff  x_{1} x_{2} x_{3} ··· x_{n} 
Probability  p(x_{1}) p(x_{2}) p(x_{3}) ··· p(x_{n}) 
then the expected value of the experiment is:
Expected Value = x_{1}p(x_{1}) + x_{2}p(x_{2}) + x_{3}p(x_{3}) + ··· + x_{n}p(x_{n})
Example 7.3.1
Number of Children  3  2  1  0 
Probability  0.10  0.60  0.20  0.10 
Example 7.3.2
Example 7.3.3
Example 7.3.4
As we have already seen, tree diagrams play an important role in solving probability problems. A tree diagram helps us not only to visualize but also to list all possible outcomes in a systematic fashion. Furthermore, when we list various outcomes of an experiment and their corresponding probabilities on a tree diagram, we gain a better understanding of when probabilities are multiplied and when they are added. The meanings of the words "and" and "or" become clear when we learn to multiply probabilities horizontally across branches, and add probabilities vertically down the tree.
Although tree diagrams are not practical in situations where the possible outcomes become large, they are a significant tool in breaking the problem down in a schematic way. We consider some examples that may seem difficult at first, but with the help of a tree diagram, they can easily be solved.
Example 7.3.5
\begin{tikzpicture}[grow=right, sloped] \tikzstyle{level 1}=[level distance=2cm, sibling distance=3cm] \node {} child { node[label=right: {$L$}] {} edge from parent node[below] {$3/4$} } child { node[label=right: {$U=1/4$}] {} edge from parent node[above] {$1/4$} }; \end{tikzpicture}  First Try 
\begin{tikzpicture}[grow=right, sloped] \tikzstyle{level 1}=[level distance=2cm, sibling distance=3cm] \node {} child { node[label=right: {$L$}] {} edge from parent node[below] {$2/3$} } child { node[label=right: {$U=(3/4)(1/3)$}] {} edge from parent node[above] {$1/3$} }; \end{tikzpicture}  Second Try 
\begin{tikzpicture}[grow=right, sloped] \tikzstyle{level 1}=[level distance=2cm, sibling distance=3cm] \node {} child { node[label=right: {$L$}] {} edge from parent node[below] {$1/2$} } child { node[label=right: {$U=(3/4)(2/3)(1/2)$}] {} edge from parent node[above] {$1/2$} }; \end{tikzpicture}  Third Try 
Example 7.3.6
Example 7.3.7
Answers
Example 5.2.1
b1s1 , b1s2 , b1s3 , b2s1 , b2s2 , b2s3
Alternatively, we can draw a tree diagram:
The tree diagram gives us all six possibilities. The method involves two steps. First the woman chooses a blouse. She has two choices: blouse one or blouse two. If she chooses blouse one, she has three skirts to match it with; skirt one, skirt two, or skirt three. Similarly if she chooses blouse two, she can match it with each of the three skirts, again. The tree diagram helps us visualize these possibilities.
The reader should note that the process involves two steps. For the first step of choosing a blouse, there are two choices, and for each choice of a blouse, there are three choices of choosing a skirt. So altogether there are 2 · 3 = 6 possibilities.
Example 5.2.2
The number of ways of choosing a blouse  The number of ways of choosing a skirt  The number of ways of choosing shoes 
2  3  2 
Example 5.2.3
Letter  Digit  Digit  Digit  Digit 
26  10  10  10  10 
Example 5.2.4
Question 1  Question 2  Question 3 
2  2  2 
Example 5.2.5
4  3  2  1 
So there are altogether 4 · 3 · 2 · 1 = 24 different ways.
Example 5.2.6
3  2  1 
Answers
Example 5.3.1
4  3  2 
Example 5.3.2
4  3 
Since there are no more restrictions, we can go ahead and make the choices for the rest of the positions. So far we have used up 2 letters, therefore, five remain. So for the next position there are five choices, for the position after that there are four choices, and so on. We get:
4  5  4  3  2  1  3 
Example 5.3.3
0! = 1
Now we define nPr.
The Number of Permutations of n Objects Taken r at a Time:
nPr = n(n − 1)(n − 2)(n − 3)···(n − r +1), or
nPr = $\frac{n!}{(nr)!}$
Where n and r are natural numbers.
The reader should become familiar with both formulas and should feel comfortable in applying either.
Example 5.3.4
Next we consider some more permutation problems to get further insight into these concepts.
Example 5.3.5
Example 5.3.6
4  3  2  5  4 
Clearly, this makes sense. For every permutation of three math books placed in the first three slots, there are 5P2 permutations of history books that can be placed in the last two slots. Hence the multiplication axiom applies, and we have the answer (4P3) (5P2). We summarize.
Example 5.3.7
1  4  3  2  1 
Example 5.3.8
1  4  3  3  2  2  1  1 
E_{1}LE_{2}ME_{3}NT
Since all the letters are now different, there are 7! different permutations. Let us now look at one such permutation, say:LE_{1}ME_{2}NE_{3}T
Suppose we form new permutations from this arrangement by only moving the E's. Clearly, there are 3! or 6 such arrangements. We list them below:LE_{1}ME_{2}NE_{3}T LE_{1}ME_{3}NE_{2}T LE_{2}ME_{1}NE_{3}T LE_{3}ME_{3}NE_{1}T LE_{3}ME_{2}NE_{1}T LE_{3}ME_{1}NE_{2}T
Because the E's are not different, there is only one arrangement LEMENET and not six. This is true for every permutation.
Let us suppose there are n different permutations of the letters ELEMENT. Then there are n · 3! permutations of the letters E_{1}LE_{2}ME_{3}NT. But we know there are 7! permutations of the letters E_{1}LE_{2}ME_{3}NT. Therefore: n · 3! = 7!
Or n = $\frac{7!}{3!}$.
This gives us the method we are looking for.
Permutations with Similar Elements:
The number of permutations of n elements taken n at a time, with r_{1} elements of one kind, r_{2} elements of another kind, and so on, is $\frac{n!}{r_1!r_2!...r_k!}$
Example 5.3.9
Example 5.3.10
Example 5.3.11
Example 5.3.12
We summarize:
$\frac{n!}{r_1!r_2!···r_k!}$
This is also referred to as ordered partitions.
Answers
Suppose we have a set of three letters {A,B,C}, and we are asked to make twoletter word sequences. We have the following six permutations:
AB BA BC CB AC CA
Now suppose we have a group of three people {A,B,C} as Al, Bob, and Chris, respectively, and we are asked to form committees of two people each. This time we have only three committees, namely:
AB BC AC
When forming committees, the order is not important, because the committee that has Al and Bob is no different than the committee that has Bob and Al. As a result, we have only three committees and not six. Forming word sequences is an example of permutations, while forming committees is an example of combinations – the topic of this section.
Permutations are those arrangements where order is important, while combinations are those arrangements where order is not significant. From now on, this is how we will tell permutations and combinations apart.
Just as the symbol nPr represents the number of permutations of n objects taken r at a time, nCr represents the number of combinations of n objects taken r at a time.
Our next goal is to determine the relationship between the number of combinations and the number of permutations in a given situation.
Example 5.4.1
Summarizing:
Example 5.4.2
Example 5.4.3
Example 5.4.4
Example 5.4.5
Example 5.4.6
Example 5.4.7
So far we have solved the basic combination problem of r objects chosen from n different objects. Now we will consider certain variations of this problem.
Example 5.4.8
2Man Committees  3Woman Committees 
M_{1}M_{2}  W_{1}W_{2}W_{3} 
M_{1}M_{3}  W_{1}W_{2}W_{4} 
M_{1}M_{4}  W_{1}W_{3}W_{4} 
M_{2}M_{3}  W_{2}W_{3}W_{4} 
M_{2}M_{4}  
M_{3}M_{4} 
Example 5.4.9
Example 5.4.10
Example 5.4.11
Ways of selecting a suit  Ways if selecting 4 cards from this suit  Ways if selecting the next suit  Ways of selecting a card from that suit 
4C1  13C4  3C1  13C1 
a. All three white
b. One of each colour
c. At least two red
Answers
$P(E^C)=\frac{nk}{n}=1\frac{k}{n}=1P(E)$
Of particular interest to us are the events whose outcomes do not overlap. We call these events mutually exclusive. Two events E and F are said to be mutually exclusive if they do not intersect. That is, $E\cap F$ =∅. Next we'll determine whether a given pair of events are mutually exclusive.Example 6.2.1
Example 6.2.2
Example 6.2.3
Example 6.2.4
Example 6.2.5
Example 6.2.6
Example 6.2.7
Example 6.2.8
Coffee drinker  Males (M)  Females (F)  TOTAL 
Yes (Y)  31  33  64 
No (N)  19  17  36 
50  50  100 
a. $P(M\cup Y)$
b. $P(F\cup N)$
5. If $P(E)=0.3$, $P(E\cup F)=0.6$, and $P(E\cap F)=0.2$, use the addition rule to find $P(F)$Answers
Example 6.3.1
Example 6.3.2
Example 6.3.3
Example 6.3.4
Example 6.3.5
Example 6.3.6
a. P (both red)
b. P (both yellow)
2. Three marbles are drawn from a jar containing five red, four white, and three blue marbles. Find the following probabilities using combinations:a. P (all three red)
b. P (none white)
3. A committee of four is selected from a total of 4 freshmen, 5 sophomores, and 6 juniors. Find the probabilities for the following events:a. At least three freshmen
b. All four of the same class
c. Exactly three of the same class
Answers
Algebraic term  Description  Example 
Algebraic expression  A mathematical phrase that contains numbers, letters, grouping symbols (parentheses) and arithmetic operations.  $5x + 2 , 3a + (4b  6) , \frac{2}{3} + 4$ 
Constant  A number.  $x + \bf2$ constant: $2$ 
Variable  A letter that can be assigned different values.  $3 – \bf x$ variable: $x$ 
Coefficient  The number in front of a variable.  $\bf6$$x$ coefficient: $6$ $x$ coefficient: $1$ 
Term  A term can be a constant, variable, or the product of a number and variable(s). (Terms are separated by addition or subtraction signs)  $3x  \frac{2}{5} + 13y^2 + 7xy$ Terms: $3x$ , $\frac{2}{5}$ , $13y^2$ , $7xy$ 
Like terms  The terms that have the same variables and exponents.  $2x  y^2 \frac{2}{5} + 5x  7 + 13y^2$ Like terms: $2x$ and $5x$ , $y^2$ and $13y^2$ , $\frac{2}{5}$ and $7$ 
Addition (+)  Subtraction ()  Multiplication (×)  Division (÷)  Equals to (=) 
add  subtract  times  divided by  equals 
sum (of)  difference  product  quotient  is 
plus  take away  multiplied by  over  was 
total (of)  minus  double  split up  are 
altogether  less (than)  twice  fit into  were 
increased by  decreased by  triple  per  amounts to 
gain (of)  loss (of)  of  each  totals 
combined  (amount) left  how much (total)  goes into  results in 
in all  savings  how many  as much as  the same as 
greater than  withdraw  out of  gives  
complete  reduced by  ratio /rate  yields  
together  fewer (than)  percent  
more (than)  how much more  share  
additional  how long  average 
Steps for solving word problems 
Exponential notation  Example 
Power Exponent  
a^{n}= a ∙ a ∙ a ∙ a … a  2^{4} = 2 ∙ 2 ∙ 2 ∙ 2 = 16 
Base  
Read “a to the nth” or “the nth power of a.”  Read “2 to the 4th.” 
Name  Property 
Zero exponent a^{0}  a^{0} = 1 (0^{0} is undefined) 
One exponent a^{1}  a^{1} = a 
1^{n} = 1 
Order of operations 
1. the brackets or parentheses (innermost first)  ( ) , [ ] , { } 
2. exponent (power)  a^{n} 
3. multiplication and division (from lefttoright)  × and ÷ 
4. addition and subtraction (from lefttoright)  + and – 
B  E  DM  AS 
Brackets  Exponents  Divide or Multiply  Add or Subtract 
 Convert a smaller unit to a larger unit: move the decimal point to the left.
 Convert a larger unit to a smaller unit: move the decimal point to the right.
Convert units using the unit factor method (or the factorlabel method):(Put the desired or unknown unit on the top.)
Prefix  Symbol (abbreviation)  Power of 10  Example 
mega  M  10^{6}  1 Mm = 1,000,000 m 
kilo  k  10^{3}  1 km = 1,000 m 
hecto  h  10^{2}  1 hm = 100 m 
deka  da  10^{1}  1 dam = 10 m 
meter/gram/liter  1  
deci  d  10^{1}  1 m = 10 dm 
centi  c  10^{2}  1 m = 100 cm 
milli  m  10^{3}  1 m = 1,000 mm 
micro  µ  10^{6}  1 m = 1,000,000 µm 
Value  1,000,000  1,000  100  10  1  .  0.1  0.01  0.001  0.000 001 
Prefix  Mega  kilo  hecto  deco  meter (m) gram (g) liter (L)  .  deci  centi  milli  micro 
Symbol  M  k  h  da  .  d  c  m  µ 
a) 439 mm = ( ? ) m
b) 236 hg = ( ? ) g
c) 3 mL = ( ? ) kL
d) 5 kg = ( ? ) hg
2. Convert each measurement using the unit factor method.a) 7230 g = ( ? ) kg
b) 52 cm = ( ? ) mm
c) 4 dL = ( ? ) L
d) 52 daL = ( ? ) cL
3. Combine.a) 7 m – 3000 mm = ( ? ) mm
b) 63 kg + 6 g = ( ? ) g
c) 72 L + 4.58 L – 10 mL = ( ? ) mL
d) 3 km + 357 dam = ( ? ) km
4. Convert.a) 7400 cm^{2} = ( ? ) m^{2}
b) 09 km^{2} = ( ? ) m^{2}
c) 5 m^{3} = ( ? ) cm^{3}
d) 567 mm^{3} = ( ? ) cm^{3}
5. Complete.a) A cube holds 1 mL of water and has a mass of 1 gram at ( ) °C.
b) 38 cm^{3} = ( ) g
c) 5 L = ( ) cm^{3}
d) 27 cm^{3} = ( ) cL
e) 76 cm^{3} of water at 4°C has a mass of ( ) g.
f) 18 L of water has a volume of ___________ cm^{3} .
g) 257 kg = ( ) L
h) A fish box that measures 45 cm by 35 cm by 25 cm. How many kiloliters of water will it hold?
Answers
a) $2x  3$
b) $4t + 13 + \frac{5}{7}t$
2. Identify the terms for each of the following:a) $5x + 3  y$
b) $2r + 16r^2 \frac{3}{14}r + 1$
3. Identify the like terms in the following expressions:a) $7 + 2y^2 \frac{5}{9}x + 5x  1 + 13y^2$
b) $0.6t + 9uv  7t + 1.67uv$
4. Evaluate the following algebraic expressions.a) $7x  4 + 13x$, given $x = 4$.
b) $\frac{3}{a7} + 9b + 12$, given $a = 10$ and $b = 5$.
5. Write an expression/equation for each of the following:a) The product of ten and y.
b) The quotient of t and six.
c) The difference between fifteen and a number more than the quotient of three by seven is six.
d) Seven less than six times a number is fifteen.
6. Write an expression for each of the following:a) Susan has \$375 in her checking account. If she makes a deposit
of y dollars, how much in total will she have in her account?
b) Mark weighs 175 pounds. If he loses y pounds, how much
will he weigh?
c) A piece of wire 45 meters long was cut in two pieces and
one piece is w meters long. How long is the other piece?
d) Emily made 4 dozen muffins. If it cost her x dollars, what was her cost per dozen muffins? What was her cost per muffin?
7. Solve.a) In $x^3$, the base is ( ).
b) In $y^4$, the exponent is ( ).
8. Write the following exponential expressions in expanded form.a) $9^3$
b) $(y)^4$
c) $(0.5a^3b)^2$
d) $(\frac{2}{7}x)^1$
9. Write each of the following in the exponential form.a) $(0.06) (0.06) (0.06) (0.06)$
b) $(12y) (12y) (12y)$
c) $(\frac{2}{9}x) (\frac{2}{9}x)$
10. Evaluate $(3^2) (2^4)(23^0)(10^1)$. 11. Write each of the following as a base with an exponent.a) y to the eighth power.
b) Five cubed.
12. Evaluate the following:a) $\frac{9a^2}{b+6} + 3a + 4$ , if $a = 1$ and $b = 3$.
b) $8xy + 7y4$ , if $x = \frac{1}{4}$ and $y = 1$.
13. Calculate the following:a) $2 \cdot 4^3 + 7  (4 + 3) + 5$
b) $5 \cdot 7 + [11 + (4 – 3)] + 4^2$
c) $\frac{1044^2}{6+5}$
Answers
Name  Additive properties  Multiplicative properties 
Commutative property  a + b = b + a  a b = b a 
Associative property  (a + b) + c = a + (b + c)  (a b) c = a (b c) 
Identity property  a + 0 = a  a ∙ 1 = a 
Closure property  If a and b are real numbers, then a + b is a real number.  If a and b are real numbers, then a ∙ b is a real number. 
Inverse property  a + a = 0  $a \cdot \frac{1}{a} = 1$ 
Distributive property  a (b + c) = ab + ac  
Property of zero  a ∙ 0 = 0 
Order of operations 
Clear the brackets or parentheses and absolute values (innermost first).  ( ) , [ ] , { } and   
Calculator exponents (power) and radicals.  a^{n} and $\sqrt$ 
Perform multiplication or division (from lefttoright).  × and ÷ 
Perform addition or subtraction (from lefttoright).  + and  
Operation  Method 
Adding signed numbers 
Add their values, and keep their common sign.
Subtract their values, and keep the sign of the larger number. 
Subtracting signed numbers  Subtract a number by adding its opposite. 
Multiplying signed numbers  (+) (+) = (+), (–) (–) = (+), (–) (+) = (–), (+) (–) = (–) 
Dividing signed numbers  $\frac{+}{+}=+$, $\frac{}{}=$, $\frac{+}{}=$, $\frac{}{+}=$ Note: $\frac{0}{A}= 0$, $\frac{A}{0}$ is undefined. 
a) 12 a + 0 = 12a
b) (3x + 11y) + 7 = 7 + (3x + 11y)
c) (4 + x) + 11 = 4 + (x + 11)
d) (6a + 5) + [(6a + 5)] = 0
e) 7(3y + 4) = 7 ∙ 3y + 7 ∙ 4
= 21y + 28
f) (0.5a) b = 0.5 (a b)
g) (4x) (7y) = (4 ∙ 7) (xy)
h) (8y) ∙ $\frac{1}{(8y)}$ = 1
i) (4 – 7y) 3 = 12 – 21y
j) $\frac{1}{23+7x}$ ∙ 0 = 0
k) (199 + 36) + 1 = (199 + 1) + 36
l) (1000 ∙ 8) ∙ 9 = 1000 (8 ∙ 9)
2. Regroup and simplify the calculations using properties.a) 12 + (45 + 88)
b) 9 (1000 ∙ 8 )
c) 3 + (2997 + 56)
3. Use the distributive property to write an equivalent expression without parentheses.a) 4y (y + 0.3)
b) (2 – 3y^{2}) 5
c) $\frac{1}{3}(\frac{2}{3}\frac{1}{2}x )$
4. Compare these numbers using either < or > .a) 6 8
b) 0 6
c) 4  2
d) $\frac{3}{7}$ $\frac{1}{7}$
e) 0.6 0.8
f) $1\frac{1}{2}$ $\frac{3}{8}$
5. Arrange the following numbers from the smallest to the largest number (using < to order them).a) 8, 9, 4, 23, 0, 17
b) 0.05 , 8 , $\frac{2}{5}$, $\frac{3}{5}$, 3.24
c) $\frac{1}{3}$, $\frac{2}{5}$, $\frac{1}{7}$, $1\frac{3}{4}$
6. Preform the indicated operation.a) 67
b) 35  14
c)  0.45 + 0
d)   7^{2}
e)  $\frac{1}{8}$
7. Preform the indicated operation.a) 4 [7  3 + (30  5)]
b) $\frac{9}{3^2}$ + (27  3)
8. Preform the indicated operation.a) 13 + 24
b) (7) + (8)
c) $\frac{1}{5}+(2\frac{2}{5})$
d) 9 + (4)
e) (25) + 12
f) 8.4 + (0.9)
g) (7) – (6)
h) (5) – (7)
i) $\frac{3}{7}\frac{2}{7}$
j) $\frac{1}{7}1\frac{3}{4}$
k) 45 $\div$ (9)
l) $\frac{3.6}{6}$
m) $\frac{9}{5} \div (\frac{1}{15})$
n) 72 $\div$ 9
o) $\frac{0}{1789}$
p) $\frac{3.78}{0}$
9. Write the additive inverse (opposite) of each number.a) – 45
b) $\frac{5}{8}$
c) 1
10. If x = 2 , y = 5, z = 4 and w = 0, evaluate each of the following.a) zy + x3
b) x^{2} – 2xy + y^{2} + $\frac{w}{3xyz}$
c) (x + y) (x – y) – 5z
d) 4 $(\frac{2xy}{3w})$
Answers
Algebraic term  Description  Example 
Algebraic expression  A mathematical phrase that contains numbers, variables (letters), and arithmetic operations (+, – , ×, ÷, etc.). 
3x – 4 5a^{2} – b + 3 
Constant  A number on its own.  2y + 5 constant: 5 
Coefficient  The number in front of a variable. 
9x^{2} coefficient: 9 x coefficient: 1 
Term  A term can be a constant, a variable, or the product of a number and variable. (Terms are separated by a plus or minus sign.) 
7a^{2} – 6b + 8 Terms: 7a^{2}, 6b, 8 
Like terms  The terms that have the same variables and exponents (differ only in their coefficients). 
2x and 7x 4y^{2} and 9y^{2} 
Polynomial  Example 
Monomial (one term)  0.67x 
Binomial (two terms)  4x – $\frac{2}{3}$ 
Trinomial (three terms)  2a^{2} – ab + 5 
Polynomial (one or more terms)  2xy, 4x^{3} + 11 , – $\frac{2x}{3}$ – 5y + 4 
Name  Rule  Example 
Product of like bases (The same base)  a^{m} a^{n} = a^{m + }^{n } (a ≠ 0)  2^{3}2^{2} = 2^{3+2} = 2^{5 }= 32 
Quotient of like bases (The same base)  $\frac{a^m}{a^n}$ = a^{m  }^{n }^{ } (a ≠ 0)  $\frac{y^3}{y^2}$ = y^{32} = y^{1 }= y 
Negative exponent  a^{}^{n} = $\frac{1}{a^n}$ (a ≠ 0)  4^{2} = $\frac{1}{4^2}$ = $\frac{1}{16}$ 
(a + b) (c + d) = ac + ad + bc + bd F O I L 
F  First terms  first term × first term (a + b) (c + d) 
O  Outer terms  outside term × outside term (a + b) (c + d) 
I  Inner terms  inside term × inside term (a + b) (c + d) 
L  Last terms  last term × last term (a + b) (c + d) 
a) 5x^{3} 8x^{2} + 2x
b) –$\frac{2}{3}$y^{4} + 9a^{2} + a – 1
2. Identify the coefficients and the degree of the polynomials.a) 2a^{3} – 7a^{2}b^{3} + 9b + 11
b) 8xy^{5} – $\frac{2}{3}$y^{4} + 11x^{2}y^{3} + 4y^{2} – 23y + $\frac{5}{6}$
3. Identify each polynomial as a monomial, binomial, or trinomial.a) 3x^{2} – 7x
b) 29xy^{3}
c) 8mn^{2} + 7m – 45
4. Arrange polynomials in descending order.a) 3 + 8x 23x^{2} + 15x^{3}
b) 3y^{3} – 45y^{2} + 4y + $\frac{2}{3}$y^{4}
5. Combine like terms.a) 7x + 10y – 8x + 9y
b) 12a^{2} – 33b + 2b – 6a^{2}
c) 12uv^{2} –5u^{2}v + 15u^{2}v – 8uv^{2}
d) 5(4t – 6r) + 3(t + 7r)
e) 13n + 5(6n – m^{2}) + 7(2m^{2} + 3n)
6. Simplify.a) 15a^{2} + 9 – (5a^{2} – 4)
b) (13x + 9y) – 6(x – 5y)
c)  (7z^{2} + 6z – 15) + 2(7z^{2} – 5z + 8)
d) 11(y^{2} – 3y) + 4(2y – 5) – (13 – 6y + 9y^{2})
e) 5(ab – 2xy) – 6(2ab + 3xy)
7. Simplify the following.a) a^{3} a^{6}
b) $\frac{x^{4}}{x^7}$
c) $\frac{t^3}{t^9}$
d) (6a^{3} b^{5}) (7a^{4} b^{6})
e) $(\frac{5}{6}x^3y^4z^5)(\frac{3}{10}xy^3z^4)$
f) $\frac{6y^8}{36y^3}$
g) $\frac{81m^3n^9}{9m^4n^9}$
8. Perform the indicated operation.a) 4x^{3} (3x^{4} – 7x)
b) 9a^{3}b (3ab^{2} + 2a^{2} b^{2} – a)
c) $\frac{35a^2+5a4}{5a}$
d) (5y – 7) (8y + 9)
e) (7r – 2t) (3r + 4t^{2})
f) (2ab^{2} + 3b) (5a^{2}b + 3a)
g) (x – $\frac{1}{3}$) (x – $\frac{2}{3}$)
Answers
Examples
3 + 2 = 5 in algebra may look like x + 2 = 5  x represents 3 
Examples
$2$ in $x + 2$ is a constant. 
Examples
$x + 2$  when $x = 0$,  $x + 2 = 0 + 2 = 2$ 
when $x = 3$,  $x + 2 = 3 + 2 = 5$ 
Examples
$9x$  coefficient:  $9$  
$\frac{2}{7}x$  coefficient:  $\frac{2}{7}$  
$x$  coefficient:  $1$  $x = 1\cdot x$ 
Examples
$5x + 2,\quad\frac{2y}{3} + 4,\quad(3x  4y^2) + 6$ 
Examples
a) $3x – 4 + \frac{2}{5} + y$  has four terms:  $3x\; , \; 4 \; , \; \frac{2}{5} \;$ and $y$. 
b) $7xyz + 12 – \frac{4}{19}z^2$  has three terms:  $7xyz\; , \; 12$ and $\frac{4}{19}z^2$. 
Examples
Like terms:  $2x$ and $5x$  The same variable: $x$ 
$– 3y^2$ and $4y^2$  The same variable raised to the same power: $y^2$  
$\frac{6}{7}$ and $9$  All constants are like terms. 
Examples
1) $3x – 4$, given $x = 5$.  
$3x – 4 = 3\cdot 5 – 4$  Substitute $x$ for $5$. 
$= 15 – 4$  Calculate. 
$= 11$ 
2) $\frac{x}{y} + 8$, given $x = 9$ and $y = 3$.  Substitute $x$ for $9$ and $y$ for $3$. 
$\frac{x}{y} + 8 = \frac{9}{3} + 8$  
$= 5$ 
3) $3a – 4 + 2$, given $a = 5$.  
$3a – 4 + 2 = 3\cdot 5 – 4 + 2$  Substitute $a$ for $5$. 
$= 15 – 4 + 2$  Calculate. 
$=13$ 
4) $\frac{6x}{y3} + 7x  2$, given $x = 1$ and $y = 9$.  
$\frac{6x}{y3} + 7x  2 = \frac{6\cdot 1}{93} + 7\cdot 1  2$  Substitute $x$ for $1$ and $y$ for $9$. 
$= \frac{6}{6} + 7  2$  Calculate. 
$= 6$ 
Addition (+)  Subtraction ()  Multiplication (×)  Division (÷)  Equals to (=) 
add  subtract  times  divided by  equals 
sum (of)  difference  product  quotient  is 
plus  take away  multiplied by  over  was 
total (of)  minus  double  split up  are 
altogether  less (than)  twice  fit into  were 
increased by  decreased by  triple  per  amounts to 
gain (of)  loss (of)  of  each  totals 
combined  (amount) left  how much (total)  goes into  results in 
in all  savings  how many  as much as  the same as 
greater than  withdraw  out of  gives  
complete  reduced by  ratio /rate  yields  
together  fewer (than)  percent  
more (than)  how much more  share  
additional  how long  average 
Examples
1) Edward drove from Prince George to Williams Lake (235 km), then to Cache Creek (203 km) and finally to Vancouver (390 km). How many kilometers in total did Edward drive?
235km + 203 km + 390 km = 828 km  The key word: total (+) 
2) Emma had \$150 in her purse on Friday. She bought a pizza for \$15, and a pair of shoes for \$35. How much money does she have left?
\$150 – 15 – 35 = \$100  The key word: left (–) 
3) Lucy received \$950 per month of rent from Mark for the months September to November. How much rent in total did she receive?
\$950 $\cdot$ 3 = \$2850  The key word: how much total (×) 
4) Julia is going to buy a \$7500 used car from her uncle. She promises to pay \$500 per month. In how many months she can pay it off?
\$7500 ÷ \$500 = 15 month  The key word: per (÷) 
Algebraic expression  Word phrases  Algebraic expression  Word phrases 
$7 + y$  the sum of 7 and y  $t  8$  8 less than t 
7 more than y  t decreased (or reduced) by 8  
y increased by 7  subtract 8 from t  
7 plus y  the difference between t and 8 
$2x$ or $2\cdot x$  the product of 2 and x  $z\div 3$ or $\frac{z}{3}$  the quotient of z and 3 
2 multiplied by x  z divided by 3  
double (or twice) of x  one third of z 
$y^3$  the third power of y  $4y  9$  9 less than 4 times y 
y cubed  $2(t  5)$  twice the difference of t and 5  
y raised to the third power  $6 + \frac{2x}{3}$  6 more than the quotient of 2x by 3 
Examples
1) The difference between t and 7 means t – 7 not 7 – t.  t appears first. 
2) 8 less than t means t – 8 not 8 – t.  8 less than t not t less than 8. 
3) The quotient of z and 3 means $\frac{z}{3}$ not $\frac{3}{z}$.  z appears first. 
Examples
1) Five greater than four divided by a number is seventeen.  Equation 
5 + 4 ÷ x = 17  $5 +\frac{4}{x} = 17$ 
(Let x = a number)  
2) A number is 7 times the number y added to 23.  
x = 7 $\cdot$ y + 23  $x = 7y + 23$ 
(Let x = a number) 
Examples
Expression  
1) The difference of y and 3.45.  $y – 3.45$ 
2) The difference of $\frac{4}{23}$ and w.  $\frac{4}{23}  w$ 
3) z less than the number 67.  $67  z$ 
4) 27 minus the product of 18 and a number. (Let x = a number)  $27  18x$ 
5) The sum of a number and 7 divided by 2. (Let x = a number)  $\frac{x + 7}{2}$ 
6) Steve has \$200 in his saving account. If he makes a deposit of x dollars, how much in total will he have in his account?  $200 + x$ 
7) Ann weighs 150 pounds. If she loses y pounds, how much will she weigh?  $150  y$ 
8) A piece of wire 30 centimeters long was cut in two pieces and one piece is z centimeters long. How long is the other piece?  $30  z$ 
9) Alice made 3 dozen cupcakes. If it costs her y dollars, what was her cost per dozen cupcakes? What was her cost per cupcake?  $\frac{y}{3} , \frac{y}{36}$ 
(1 dozen = 12 , 3 $\cdot$ 12 = 36) 
Steps for solving word problems 
 Organize the facts given from the problem (create a table or diagram if it will make the problem clearer)
 Identify and label the unknown quantity (let x = unknown).
 Convert words into mathematical symbols, and determine the operation – write an equation (looking for ‘key’ or ‘clue’ words).
 Estimate and solve the equation and find the solution(s).
 Check and state the answer.
(Check the solution to the equation and check it back into the problem – is it logical?) 
Examples
5 socks  \$4.35 each 
Sales tax  \$2.15 
Money left  \$5.15 

\$4.35 × 5 

(\$4.35 × 5) + \$2.15 

x = [(\$4.35 × 5) + \$2.15] + \$5.15 

x = [(\$4 × 5) + \$2] + \$5 
= \$27  

x = [(\$4.35 × 5) + \$2.15] + \$5.15 
= \$29.05 
\$29.05  [(\$4.35 × 5) + \$2.15] $\overset{?}{=}$ \$5.15  
\$29.05 – \$23.9 $\overset{\checkmark}{=}$ \$5.15  Correct! 
Examples
James had  96 toys 
The total number of toys sold  13 + 32 + 21 + 14 + 7 
The toys not sold  96 – the total number of toys sold 

Let x = percentage of the toys were not sold 

13 + 32 + 21 + 14 + 7 = 87 

96 – 87 = 9 

x = $\frac{Toys\,not\,sold}{Total\,number\,of\,toys}$ = $\frac{9}{96} \approx$ 0.094 = 9.4% 

9.4% percentage of the toys were not sold 
Examples
60 L × = 30 L  Robert has 30 liters gas in his car. 
$\frac{7L}{100 km} = \frac{x}{390 km}$  Proportion: $\frac{a}{b} = \frac{c}{d}$ 
(x)(100km) = (7 L) (390 km)  Cross multiply and solve for x. 
x = $\frac{(7 L)(390 km}{100 km}$ = 27.3 L  Robert needs 27.3 liters gas to get to Vancouver. 
Exponential notation  Example 
Power Exponent  
a^{n}= a ∙ a ∙ a ∙ a … a  2^{4} = 2 ∙ 2 ∙ 2 ∙ 2 = 16 
Base  
Read “a to the nth” or “the nth power of a.”  Read “2 to the 4th.” 
Examples
Examples
Anything raised to the first power is itself.  (4 is multiplied by itself one time) 
Examples
Name  Property  Example 
Zero exponent a^{0}  a^{0} = 1 (0^{0} is undefined)  $(\frac{3}{4})^0 = 1$ , $(2xy)^0 = 1$ 
One exponent a^{1}  a^{1} = a  $4.5^1 = 4.5$ , $(3x)^1 = 3x$ 
1^{n} = 1  $1^7 = 1$ , $1^{389} = 1$ 
Base Exponent  Repeated multiplication  Product  Read 
3^{2}  3 ∙ 3  9  3^{2}  3 squared 
10^{3}  10 ∙ 10 ∙ 10  1000  10^{3}  10 cubed 
(0.2)^{2}  0.2 ∙ 0.2  0.04  (0.2)^{2}  0.2 squared 
1^{10}  1 ∙ 1 ∙ 1 ∙ 1∙ 1 ∙ 1 ∙ 1 ∙ 1 ∙ 1 ∙ 1  1  1^{10}  1 to the tenth 
$\bf{(\frac{2}{3})^3}$  $\frac{2}{3}$ ∙ $\frac{2}{3}$ ∙ $\frac{2}{3}$  $\frac{8}{27}$  $(\frac{2}{3})^3$  two thirds cubed 
10000^{0}  1  10000^{0}  10000 to the zero  
y^{5}  y ∙ y ∙ y ∙ y ∙ y  y^{5}  y^{5}  y to the fifth 
Examples
Exponential expressions  Expanded form  
1)  $6^4$  $6$ ∙ $6$ ∙ $6$ ∙ $6$  a^{n} = a ∙ a ∙ a … 
2)  $(x)^3$  $(x)\,(x)\,(x)$  
3)  $(3x^2y)^2$  $(3x^2y)\,(3x^2y)$  
4)  $(\frac{3}{4}u)^4$  $(\frac{3}{4}u)\,(\frac{3}{4}u)\,(\frac{3}{4}u)\,(\frac{3}{4}u)$ 
Examples
Expanded form  Exponential notation  
1)  $(0.2)\,(0.2)\,(0.2)$  $(0.2)^3$ 
2)  $(5a)\,(5a)\,(5a)\,(5a)$  $(5a)^4$ 
3)  $(\frac{5}{7}t)\,(\frac{5}{7}t)$  $(\frac{5}{7}t)^2$ 
Examples
4^{2} ∙ 3^{3} ∙ 6^{0} ∙ 9^{1} = (4 ∙ 4) (3 ∙ 3 ∙ 3) (1) (9)  a^{0} = 1 , a^{1} = a 
= 16 ∙ 27 ∙ 1 ∙ 9 = 3888 
Examples
1)  Six to the power of eight.  6^{8} 
2)  x to the seventh power.  x^{7} 
3)  Eight cubed.  8^{3} 
Examples
$\frac{6x^2}{y 3} + 7x  2 = \frac{6\cdot 2^2}{9 3} + 7\cdot 2  2$  Substitute x for 2 and y for 9. 
$=\frac{24}{12} + 14 2 = 14$  Calculate. 
Order of operations 
1. the brackets or parentheses (innermost first)  ( ) , [ ] , { } 
2. exponent (power)  a^{n} 
3. multiplication and division (from lefttoright)  × and ÷ 
4. addition and subtraction (from lefttoright)  + and – 
Examples
4 ∙ 3^{2} + 5 (2 + 1) – 2 = 4 ∙ 3^{2} 5 + 3 – 2  ( ), a^{n} 
= 4 ∙ 9 + 5 + 3 – 2  × 
= 36 + 5 + 3 – 2  + 
= 41 + 3 – 2  + 
= 44 – 2  – 
= 42 
B  E  DM  AS 
Brackets  Exponents  Divide or Multiply  Add or Subtract 
Examples
4 ∙ 3 [5 + (2 + 1)] – 3^{2} = 4 ∙ 3 [5 + 3] – 3^{2}  ( ), [ ] 
= 4 ∙ 3 + 8 – 3^{2}  a^{n} 
= 4 ∙ 3 + 8 – 9  × 
= 12 + 8 – 9  + 
= 20 – 9  – 
= 11 
Additive Properties  Example 
Commutative property (switch order)  a + b = b + a  2 + 3 = 3 + 2 
Associative property (switch parentheses)  (a + b) + c = a + (b + c)  (2 + 1) + 3 = 2 + (1 + 3) 
Identity property  a + 0 = a  100 + 0 = 100 
Closure property  If a and b are real numbers, then a + b is a real number.  2 and 5 are real numbers, so 2 + 5 = 7 is a real number. 
Inverse property  a + a = 0  2 + 2 = 0 
Examples
Answer  
1) 7 x + 0 = 7x  Identity property 
2) (97 + 22) + 3 = (97 + 3) + 22  Commutative property (switch order) 
3) (3 + 11x) + 7x = 3 + (11x + 7x)  Associative property (switch parentheses) 
4) (4y + 3) + [(4y + 3)] = 0  Inverse property of addition 
Multiplicative properties  Example 
Commutative property (Switch order)  a b = b a  2 ∙ 3 = 3 ∙ 2 
Associative property (Switch parentheses)  (a b) c = a (b c)  (2 ∙ 1) 3 = 2 (1 ∙ 3) 
Identity property of 1  a ∙ 1 = a  100 ∙ 1 = 100 
Closure property  If a and b are real numbers, then ab is a real number.  3 and 4 are real numbers, so 3 (4) = 12 is a real number 
Distributive property  a (b + c) = ab + ac a (b – c) = ab – ac  2 (3 + 4) = 2 ∙ 3 + 2 ∙ 4 3 (4 – 2) = 3 ∙ 4 – 3 ∙ 2 
Property of zero  a ∙ 0 = 0  35 ∙ 0 = 0 
Inverse property  $a \cdot \frac{1}{a} = 1$  $5 \cdot \frac{1}{5} = 1$ 
Examples
Answer  
1) (3y) (5y) = (5 ∙ 3) (y ∙ y) = 15y^{2} 
Commutative property of multiplication 
2) (9x) x^{2 }= 9 (x ∙ x^{2}) = 9x^{3} 
Associative property of multiplication 
3) $\frac{1}{5}$ (10 x – 15) = $\frac{1}{5}$ ∙ 10x – $\frac{1}{5}$ ∙ 15 = 2x – 3 
Distributive property of multiplication 
4) –(7 + 3x) ∙ $\frac{1}{(7+3x)}$ = 1 
Inverse property of multiplication 
5) (2x – 3y) x = 2x^{2} – 3xy 
Distributive property 
6) $\frac{1}{4x}$ ∙ 0 = 0 
Multiplicative property of zero 
7) (1000 ∙ 8) ∙ 9 = 1000 (8 ∙ 9) = 1000(72) = 72000 
Associative property of multiplication 
Name  Additive properties  Multiplicative properties 
Commutative property  a + b = b + a  a b = b a 
Associative property  (a + b) + c = a + (b + c)  (a b) c = a (b c) 
Identity property  a + 0 = a  a ∙ 1 = a 
Closure property  If a and b are real numbers, then a + b is a real number.  If a and b are real numbers, then a ∙ b is a real number. 
Inverse property  a + a = 0  $a \cdot \frac{1}{a} = 1$ 
Distributive property  a (b + c) = ab + ac  
Property of zero  a ∙ 0 = 0 
Examples
1) (43 + 1998) + 2 = ?
43 + (1998 + 2) = 2043 
Associative property of addition 
2) (7 ∙ 1000) 9 = ?
(7 ∙ 9) ∙ 1000 = 63,000 
Commutative property of multiplication 
Examples
1) 3 (4 + 2) = ?  
a) 3 ∙ 6 = 18 

b) 3 ∙ 4 + 3 ∙ 2 = 18 
Distributive property 
2) $\frac{1}{2}(\frac{1}{2} + 1\frac{2}{3})$ = ?  
a) $\frac{1}{2}(\frac{1}{2} + \frac{5}{3}) = \frac{1}{2}(\frac{3}{6} + \frac{10}{6})$ = $\frac{1}{2}(\frac{13}{6}) = \frac{13}{12} = 1\frac{1}{12}$ 
$1\frac{2}{3} = \frac{5}{3}$ 
b) $\frac{1}{2}(\frac{1}{2} + \frac{5}{3}) = \frac{1}{2}(\frac{1}{2}) + \frac{1}{2}(\frac{5}{3})$ = $\frac{1}{4}+\frac{5}{6} = \frac{3}{12}+\frac{10}{12} = \frac{13}{12} = 1\frac{1}{12}$ 
Distributive property 
Negative number: 3, 2 , 9x
Meaning  Example 
Temperature  + 0C: above 0 degree  0C: below 0 degree  +200C 50C 
Money  + \$: gain or own  \$: loss or owe  Own: +\$10000 Owe: \$500 
Sports  + points: gain  points: loss  Gain 3 points: +3 Lost 2 points: 2 
Examples
big > small, small < big
Examples
17 < 6 < 3 < 0 < 3 < 11
b) $\frac{1}{2}$ , $\frac{2}{3}$ , $\frac{1}{4}$ , $2\frac{2}{3}$$\frac{1}{2}$ = 0.5 , $\frac{2}{3} \approx$ 0.67 , $\frac{1}{4}$ = 0.25 , $2\frac{2}{3}$ = $\frac{8}{3} \approx$ 2.67
0.5 < 0.25 < 0.67 < 2.67
$\frac{1}{2}$ < $\frac{1}{4}$ < $\frac{2}{3}$ < $2\frac{2}{3}$
18 is 18 units away from 0.
Examples
Order of operations 
Clear the brackets or parentheses and absolute values (innermost first).  ( ) , [ ] , { } and   
Calculator exponents (power) and radicals.  a^{n} and $\sqrt{\;\;}$ 
Perform multiplication or division (from lefttoright).  × and ÷ 
Perform addition or subtraction (from lefttoright).  + and  
Examples
1) 3 [7  4 + (10  2)] = 3 [7  4 + 8] = 3 [3 + 8] = 3 ∙ 11 = 33  Parentheses 
Brackets / subtraction  
Brackets /addition  
Multiplication 
2) $\frac{8}{2^2}$  (4  3) = $\frac{8}{2^2}$  1 = $\frac{8}{4}$  1 = 2  1 =1  Parentheses and absolute value 
Exponent  
Division  
Subtraction 
Examples
1) 5 + 4 = 9  Add and keep the ( + ) sign. 
2) (6) + (2) = 8  Add and keep the (  ) sign. 
3) $\frac{1}{2}+(1\frac{1}{2}) = \frac{1}{2}+(\frac{3}{2}) = \frac{4}{2} = 2$  Add and keep the (  ) sign. 
Examples
1) 2 + (5) = 3  Subtract and keep the sign of 5, since 5 > 2. 
2) (3) + 7 = 4  Subtract and keep the sign of 7, since 7 > 3. 
3) 3.2 + (0.2) = 3  Subtract and keep the sign of 3.2, since 3.2 > 0.2. 
(Change the sign of b and then follow the rules for adding signed numbers.)
Examples
1) (3) – (4) = (3) + (4) = 1  Change the sign of the (4), then add (3) and 4. 
2) (7) – 2 = (7) + (2) = 9
– (+2) 
Change the sign of the 2, then add (7) and (2). 
3) $\frac{1}{3}\frac{2}{3} = \frac{1}{3}+(\frac{2}{3}) = \frac{3}{3} = 1$
$(+\frac{2}{3})$ 4) $\frac{3}{5}1\frac{1}{2} = \frac{3}{5}\frac{3}{2} = \frac{6}{10}\frac{15}{10} = \frac{9}{10} = \frac{9}{10}$ 
Examples
1) The additive inverse of 7 is 7  7 + (7) = 0 
2) The additive inverse of $\frac{2}{5}$ is $\frac{2}{5}$  $\frac{2}{5} + \frac{2}{5} = 0$ 
(3) (5) = 15
(0.3) (3) =  0.9
(4)^{2} = (4) (4) = 16
42 = 1 ∙ 4^{2} = 16
Multiplication  Example 
Positive × Positive = Positive  (+) (+) = (+)  4 ∙ 3 = 12 
Negative × Positive = Negative  (–) (+) = (–)  (4) (3) = –12 
Positive × Negative = Negative  (+) (–) = (–)  (4) (–3) = –12 
Negative × Negative = Positive  (–) (–) = (+)  (–4) (–3) = 12 
Examples
Multiplying  Example 

(3) (4) = 12 

(0.5) (0.6) = 0.3 

(4) (2) (5) (1) = 40 (1)^{4} = 1 

$(\frac{2}{3})(\frac{1}{2})(\frac{3}{4}) = \frac{1}{4}$ $(1)^7 = 1$ 
Examples
a^{4} – b + c = (1)^{4} – (2) + 4
= 1 + 2 + 4 = 7 
Substitute a for 1, b for 2 (add parentheses), and c for 4. 
2) $\frac{1.8}{2} = 0.9$
3) $\frac{8}{4} \div (\frac{1}{4}) = \frac{8}{4} \times (\frac{4}{1}) = 8$
2) $\frac{49}{7} = 7$
3) $\frac{3}{9} \div (\frac{6}{3}) = \frac{3}{9} \times (\frac{3}{6}) = \frac{1}{3} \times (\frac{1}{2}) = \frac{1}{6}$
Division  Sign  Example 
Positive ÷ Positive = Positive  $\frac{+}{+}=+$  $\frac{28}{7}=4$ 
Negative ÷ Positive = Negative  $\frac{}{+}=$  $\frac{9}{3}=3$ 
Positive ÷ Negative = Negative  $\frac{+}{}=$  $\frac{4.9}{0.7}=7$ 
Negative ÷ Negative = Positive  $\frac{}{}=$  $\frac{72}{8}=9$ 
Examples
$a^2\frac{a}{abc}+\frac{d}{c} = (2)^2\frac{2}{(2)(1)(1)}+\frac{0}{1}$
Substitute a for 2, b for 1, c for 1 and d for 0.
= $4\frac{2}{2}+0$
= 5
Properties  Equality  Example 
Addition property of equality  A = B A + C = B + C  Solve $x6=3$ $x\bcancel{6}+\bcancel{\bf6}=3+\bf6$ x = 9 
Subtraction property of equality  A = B A – C = B – C  Solve $y+5=8$ $y+\bcancel{5}\bcancel{\bf5}=8\bf5$ y = 13 
Multiplication property of equality  A = B A · C = B · C  Solve $\frac{m}{9}=2$ $\bcancel{\bf9} \cdot \frac{m}{\bcancel{9}}=2 \cdot \bf9$ m = 18 
Division property of equality  A = B $\frac{A}{C}=\frac{B}{C}$ (C ≠ 0)  Solve $3n=15$ $\frac{\bcancel{3}n}{\bcancel{\bf3}}=\frac{15}{\bf3}$ n = 5 
Equation solving strategy 

a) 2: 9x – 7 = 11
b) 17: $\frac{15}{17}$y = 9
c) $\frac{2}{3}$: 9m = 6
2. Solve the following equations.a) x – 7 = 12
b) y + $\frac{3}{8}$ = $\frac{5}{8}$
c) m – 6 = 17
d) 9t = 72
e) $\frac{3x}{2}$ = $\frac{9}{16}$
f) $\frac{y}{13}$ =  4
g)  21 + x = 7
h) y + $\frac{4}{9}$ = $\frac{3}{9}$
i) $\frac{4}{14}$x =  2
j) 19 t = 38
k) 0.8y = 0.64
l) x – $4\frac{2}{3}$ = $3\frac{2}{9}$
3. Solve the following equations.a) 14t + 5 = 8
b) 7m – 23 = 40
c) 7(x – 3) + 3x – 5 = 2(5 – 4x)
d) $\frac{1}{7}$(y + 12) = 4y – $\frac{3}{7}$y
e) 0.63x – 0.29 =  3.56x
f) 0.5t + 0.05 = 0.025
g) $\frac{x}{4}+\frac{2}{5}=\frac{x}{2}\frac{1}{5}$
Answers
Representation  Example 
Ratio  a to b or a:b or $\frac{a}{b}$  with the same unit.  5 to 9 or 5:9 or $\frac{5\;km}{9\;km}$ 
Rate  a to b or a:b or $\frac{a}{b}$  with different units.  3 to 7 or 3:7 or $\frac{3\;cm}{7\;km}$ 
Proportion  $\frac{a}{b}=\frac{c}{d}$  an equation with a ratio/rate on each side.  $\frac{x\;m}{5\;km}=\frac{3\;m}{8\;km}$ , $\frac{x\;m}{7\;m}=\frac{2\;m}{5\;m}$ 
Note: the units for both numerators must match and the units for both denominators must match. 
Conversion  Steps  Example 
Percent ⇒ Decimal  Move the decimal point two places to the left, then remove %.  31% = 31.% = 0.31 
Decimal ⇒ Percent  Move the decimal point two places to the right, then insert %.  0.317 = 0. 317 = 31.7 % 
Percent ⇒ Fraction  Remove %, divide by 100, then simplify.  15% = $\frac{15}{100}=\frac{3}{20}$ 
Fraction ⇒ Percent  Divide, move the decimal point two places to the right, then insert %.  $\frac{1}{4}$ = 1÷4 = 0.25 = 25 % 
Decimal ⇒ Fraction  Convert the decimal to a percent, then convert the percent to a fraction.  0.35 = 35 % = $\frac{35}{100}=\frac{7}{20}$ 