Mathematics 2204
Final Exam Review
Topic 1: Solving a system of equations in three variables.
2009
Solve the following system of equations using either substitution or elimination:
𝑥 + 𝑦 + 2𝑧 = 0
3𝑥 + 2𝑦 − 𝑧 = 13
𝑥 − 2𝑦 + 𝑧 = −1
Select 2 equations and eliminate one of the variables.
2nd and 3rd equations can be added to eliminate z and y.
3𝑥 + 2𝑦 − 𝑧 = 13
𝑥 − 2𝑦 + 𝑧 = −1
4𝑥
= 12
𝑥=3
Substitute this 3 for x in the 1st and 2nd or 3rd equations:
3 + 𝑦 + 2𝑧 = 0
𝑦 + 2𝑧 = −3
1st equation
3 − 2𝑦 + 𝑧 = −1
−2𝑦 + 𝑧 = −4
2nd equation
𝑦 + 2𝑧 = −3 multiply this equation by 2
−2𝑦 + 𝑧 = −4
Now solve the system of two equations in two variables:
2𝑦 + 4𝑧 = −6
−2𝑦 + 𝑧 = −4 add the equations
5𝑧 = −10
𝑧 = −2
3 + 𝑦 + 2 −2 = 0 1st equation
3+𝑦−4= 0
𝑦 =0−3+4
𝑦=1
Substitute x = 3 and z = -2 in one of the original equations
to find y.
Solution: (x,y,z)
(3, 1, -2)
2008
Algebraically solve the following system of equations using substitution or elimination:
x  y  z  1
5 x  6 y  3z  13
2 x  2 y  z  5
2007
Solve the following system of equations using either substitution or elimination:
−3𝑥 + 5𝑦 + 𝑧 = −20
2𝑥 − 2𝑦 + 3𝑧 = 18
4𝑥 + 2𝑦 + 3𝑧 = 18
Topic 2: Solve a system of equations in two variables using matrices.
2009
Barry’s Bakery sells cookies and muffins. One customer orders 5 muffins and 8 cookies, and is charged
$17.50. Another customer orders 3 muffins and 5 cookies, and is charged $10.75. Create a system of
equations to represent this information and, using matrices, find the price of each item.
Solution: Let x = price of muffins and y = price of cookies
5 8
=
3 5
5𝑥 + 8𝑦 = 17.50
3𝑥 + 5𝑦 = 10.75
Matrix equation:
𝐴
17.50
=
10.75
𝐵
𝐴 × 𝑋 = 𝐵
𝑋 = 𝐴
−1
× 𝐵
𝑥
5 8
17.50
× 𝑦 =
3 5
10.75
𝑥
5 8
𝑦 = 3 5
−1
×
17.50
10.75
𝑥
𝑦 =
2008
Lorraine buys 6 cheap golf balls and 4 expensive ones for $12.50. Bob buys 4 cheap and 3 expensive balls for $9.00.
Create a system of equations to represent this information and, using matrices, find the price of the two kinds of
balls.
2007
The student council is having a dance as a fundraiser. They are selling two kinds of tickets: singles and
couples. On the first day, 39 singles tickets and 24 couples tickets are sold for a total of $268.50. On the
second day, 13 singles tickets and 18 couples tickets are sold for a total of $144.50. Create a system of
equations to represent this information and, using matrices, find the cost for each type of ticket.
Topic 3: Finding a quadratic equation using a system of equations.
2009
Amy kicked a soccer ball. After the ball travelled a distance of 8 metres, it was at a height of 3.2 metres.
When the ball hit the ground, it was 40 metres away from Amy. Set up and solve a system of equations to
determine the quadratic function that models the path of the soccer ball.
Solution: Use the three points to find an equation of
the form 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 (Find a, b, and c.)
(x,y)
(0,0)
0=𝑎 0 2+𝑏 0 +𝑐
0= 0 + 0 +𝑐
0=𝑐
(8,3.2)
3.2 = 𝑎 8 2 + 𝑏 8 + 0
3.2 = 64𝑎 + 8𝑏
one equation
(40,0)
0 = 𝑎 40 2 + 𝑏 40 + 0
0 = 1600𝑎 + 40𝑏 second equation
Solve the system
3.2 = 64𝑎 + 8𝑏
0 = 1600𝑎 + 40𝑏
You can use any method you like (ie. substitution, elimination, matrices)
Elimination: Multiply first equation by -5 and add to second equation.
−16 = −320𝑎 − 40𝑏
0 = 1600𝑎 + 40𝑏
−16 = 1280𝑎
−16
= 𝑎 = −.0125
1280
0 = 1600 −.0125 + 40𝑏
0 = −20 + 40𝑏
20 = 40𝑏
20
= 𝑏 = .5
40
a = -.0125, b = .5, c = 0 substitute these numbers in 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
Equation:
𝑦 = −.0125𝑥 2 + .5𝑥 + 0
or
𝑦 = −.0125𝑥 2 + .5𝑥
2008
A toy rocket is launched from a platform 5 metres above the ground and reaches a height of 9 m in 4
seconds. The rocket touches down on the ground after 10 seconds. Write an equation in the form
h(t )  at 2  bt  c that describes the path of the rocket where t represents time in seconds and h(t )
represent the height of the rocket in metres.
Height
(4, 9)
(0, 5)
Tim e
(10, 0)
2007
An arrow is fired into the air from a height of 1 metre. After 2 seconds, the arrow is 9 metres in the air. After 4
seconds, the arrow is 13 metres in the air. Set up and solve a system of equations to determine the quadratic
relation, of the form h(t) at 2 bt c , that models this situation. Use this relation to determine the height,
h(t) , of the arrow at t=10 seconds.
Topic 4: Graphing transformations of y = cos x and y = sin x.
2009
Graph the following relation:
1
(3 𝑥 + 90, −2𝑦 − 1)
Solution: Find the mapping rule: (x,y)
Make two tables.
y = sin x
x
0
y
0
90
1
180
0
270
-1
360
0
1
𝑥 + 90
3
−2𝑦 − 1
90
-1
120
-3
150
-1
180
1
210
-1
2008
Graph the following relation: ( y  2)  sin
Solution: Find the mapping rule: : (x,y)
Make two tables.
y = sin x
x
0
y
0
2𝑥 − 90
−𝑦 − 2
-90
-2
90
1
90
-3
180
0
270
-2
270
-1
450
-1
360
0
630
-2
1
 x  90
2
(2𝑥 − 90, −𝑦 − 2)
Try:
Graph the following relation: 
1
 y  2   cos 2  x  90 .
4
Topic 5: Finding equations of cos or sin.
2009
A mother puts her child on a Merry-Go-Round. To watch her child, she stands at a point that is initially 3 metres away
from the child, which is the closest distance between the mother and child. At 6 seconds, the child is 15 metres from his
mother, which is the farthest distance between the mother and child. Assuming the distance between the mother and
child varies sinusoidally with time, determine the relation that models this situation.
Solution:
Find the following information:
amplitude =
𝑚𝑎𝑥 −𝑚𝑖𝑛
2
sinusoidal axis: 𝑦
=
=
2
𝑚𝑎𝑥 +𝑚𝑖𝑛
2
𝑦 = 9
Horizontal stretch:
15−3
𝑝𝑒𝑟𝑖𝑜𝑑
360
=
=
=
12
2
=6
15+3
18−6
360
2
=
distance (m)
=
12
360

18
2
=
=9



1
30


Horizontal Translation:
For cosine it is 6 or 18 right.
For sine it is 3 or 15 or 27 right.


Mapping rule: :
(x, y)
(ax ± b, cy ± d)

Horizontal stretch = a
Horizontal Translation = b
Vertical Stretch (amplitude) = c
Vertical Translation (sinusoidal axis): y = d
Reflection if c<0
Cosine:
𝑥, 𝑦 →
Equation:
1
𝑥
30
1
6
𝑦 − 9 = 𝑐𝑜𝑠30(𝑥 − 6)


time (seconds)
OR
+ 6, 6𝑦 + 9

Sine:
1
𝑥, 𝑦 → (30 𝑥 + 3,6𝑦 + 9)
Equation:
1
6
𝑦 − 9 = 𝑠𝑖𝑛30(𝑥 − 3)
Try:
Jack is riding a Ferris wheel. The graph below shows his height above the ground with relation to time. Determine
the equation of the function in terms of sine or cosine that describes Jack’s height above the ground in relation to
time.

height(m)









time(s)





Topic 6: Simplifying rational expressions
2009
Simplify the following expression:
1
2
 2
x  2 x  2x
Solution:
Factor where possible
Find the common denominator
1
𝑥+2
=
+
2
𝑥 𝑥+2
𝑥∙1
+ 2
𝑥(𝑥+2) 𝑥(𝑥+2)
=
=
𝑥+2
𝑥(𝑥+2)
1
𝑥
the (x+2)s cancel out






2008
x2  4
x2  5x
Simplify the expression :
Factor where possible
=
Change to multiplication and write the reciprocal
=

x2  4x  4
x 2  7 x  10
𝑥+2 (𝑥−2)
(𝑥+2)
÷ 𝑥+2
𝑥(𝑥+5)
𝑥+5 (𝑥+2)
𝑥+2 (𝑥−2)
(𝑥+2)
× 𝑥+5
𝑥(𝑥+5)
𝑥+2 (𝑥+2)
= (𝑥−2)
𝑥
Cancel out like terms
Try:
1. Simplify the following expression.
2. Simplify the following expression.
𝑥 2 +5𝑥+6
5
3

x 1 x  2
𝑥 2 −9
÷
𝑥 2 +6𝑥+8
𝑥 2 −7𝑥+12
Topic 7: Trigonometric Proofs
2009


Prove: sec 1  sin 2   cos 
Solution:
Left Side
sec 𝜃(1 − 𝑠𝑖𝑛2 𝜃)
Right Side
=
=
cos 𝜃
𝑐𝑜𝑠 2 𝜃
cos 𝜃
=
cos 𝜃
cos 𝜃
=
cos 𝜃
1
(𝑐𝑜𝑠 2 𝜃)
cos 𝜃
cos 𝜃
2008
Prove : sin  
cos 2 
 csc 
sin 
Solution:
Left Side
sin θ
1
+
𝑐𝑜𝑠 2 𝜃
sin 𝜃
Right Side
=
csc 𝜃
𝑠𝑖𝑛2 𝜃 𝑐𝑜𝑠 2 𝜃
+
sin 𝜃
sin 𝜃
=
1
sin 𝜃
𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 2 𝜃
sin 𝜃
=
1
sin 𝜃
1
sin 𝜃
=
1
sin 𝜃
Try:
1. Prove: sec2   sec2  sin 2   1
2. Prove that
cot 𝑥 sec 𝑥 sin 𝑥
cos 𝑥
= sec 𝑥
Topic 8: Simplifying trigonometric expressions using exact values.
2009
Simplify the following expression, expressing your answer in EXACT simplest form:
sin120 cos 45  cos 240 sin 90
Solution:
3
2
2
2
6
=
=
=
4
6
4
6
−
1
2
−
−1
+
1
2
2
+
4
−1
2
4
6+2
4
=
2008
Find the EXACT value for the following expression :
tan(60)  cot(60)
Solution:
3
+
1
=
=
3 3
3
+
3
3
3
3
4 3
3
Try:
1. Simplify the following expression, expressing
your answer in EXACT simplest form:
3sin  225 cos 135  cos 330
2. Simplify, expressing your answer in exact
radical :
cos 45°
cos 30°
+
sin 45°
sin 30°
Topic 9: Solving trigonometric equations.
2009
Solve  2sin   1 sin   1  0 where 0    360 .
Solution:
Set the two factors equal to 0 and solve for 𝜃.
2 sin 𝜃 − 1 = 0
sin 𝜃 − 1 = 0
2 sin 𝜃 = 1
1
sin 𝜃 = 2
sin 𝜃 = 1
1
2
You can use your calculator to help here but you should know when sin 𝜃 = and sin 𝜃 = 1.
Using a calculator 𝜃 = 𝑠𝑖𝑛−1
𝜃 = 30°
1
2
and
𝜃 = 𝑠𝑖𝑛−1 1
𝜃 = 90°
1
0° ≤ 𝜃 < 360° means you must find all values of 𝜃 between 0° and 360° that make sin 𝜃 = 2 and sin 𝜃 = 1.
There is one more answer for sin 𝜃 =
1
2
1
. Sin is also equal to 2 in the 2nd quadrant at 𝜃 = 150°.
So there are three answers: 𝜃 = 30°, 150°, 𝑎𝑛𝑑 90°
1
1
2
2
You can use your calculator to check these: sin 30° = , sin 150° = , sin 90° = 1.
2008
Solve 2 cos   3  0 where 0    360
Solution:
Solve for 𝜃.
2 cos 𝜃 + 3 = 0
2 cos 𝜃 = − 3
cos 𝜃 =
You can use a calculator to find one answer :
− 3
2
𝜃 = 𝑐𝑜𝑠 −1
𝜃 = 150°
Cos is also equal to
− 3
2
− 3
2
in the 3rd quadrant at 𝜃 = 330°.
You can use your calculator to check theses: cos 150° = −.866025 and cos 330° = −.866025
− 3
2
= −.866025
Try:
1. Solve 2sin x  2  0 where 0  x  360 .
2. Solve the following equation for x: 4cos x 5 3 . 0  x  360
Topic 10: Confidence Intervals
2009
Good Day Tires claims that their new line of tires will last for 80000 km. A consumer research group decides to test this
claim. The group randomly selects 400 tires and tests each tire. The data from this sample shows that the mean life of a
tire is 78000 km, with a standard deviation of 1500 km.
a) Algebraically determine the 95% confidence interval for the mean life of a tire.
Solution:
Formula for 95% confidence interval is
x z
Sx
n
where z = 1.96
x = sample mean = 78000
S x = sample standard deviation = 1500
78000 − 1.96
1500
400
= 77853
n
78000 + 1.96
1500
400
= 78147
= sample size = 400
The 95% confidence interval is: 77853 to 78147
b) Does the confidence interval support the claim of Good Day Tires? Explain your answer.
Solution:
The confidence interval does not support the claim of Good Day Tires because the confidence interval does not capture
the claim of 80000 km.
2008
A factory has 2200 workers with a mean weekly salary of $650. A simple random sample of 250 of these workers is
selected. The sample has a mean weekly salary of $630 with a standard deviation $40. Algebraically calculate the 95%
confidence interval.
Solution:
Formula for 95% confidence interval is
x z
Sx
where z = 1.96
n
x = sample mean = 630
S x = sample standard deviation = 40
630 − 1.96
40
250
= 625.0415
n = sample size = 250
630 + 1.96
40
250
= 634.9585
The 95% confidence interval is: 625.0415 to 634.9585
Try:
A soft drink manufacturer advertises that a certain size of its diet cola has a mean weight of 591 grams. A random
sample of 50 bottles is selected. The sample mean is 589.3 grams and the sample standard deviation is 13.7 grams.
Algebraically determine the 95% confidence interval. Does the confidence interval support the company’s claim?
Topic 11: The Normal Distribution (Bell Curve)
2009
The number of chocolate chips in a certain brand of chocolate chip cookies is known to be normally distributed with a
mean of 15 chips per cookie and a standard deviation of 1.5. Draw and label a normal distribution curve for this
situation, and determine the percentage of cookies has between 12 and 16.5 chocolate chips.
Solution:
Sketch and label the normal. curve.
Between 12 and 16.5: 13.5% + 34% + 34% = 81.5%
2008
The speed limit for vehicles crossing the Pinware River bridge is 60 km/h. The speed of 20 000 vehicles crossing the
bridge is normally distributed with an average speed of 68 km/h and a standard deviation of 8 km/h. Draw and label a
normal distribution curve and determine what percent of the vehicles are travelling over the speed limit of 60 km/h.
Solution:
Sketch and label the normal.
Over 60 km/h: 34% + 34% + 13.5% + 2.5% = 84%
Try:
The life expectancy of a certain brand of hair dryer is known to be normally distributed with a mean of 3 years and a
standard deviation of 0.75 years. A company produces 3000 of these hair dryers. Draw and label a normal distribution
curve and determine how many hair dryers are expected to last more than 1.5 years.
Topic 12: Law of Sines, Law of Cosines, Area of a Triangle, and Right Triangle Trigonometry
2009
Algebraically determine the measure of  given that  is obtuse.
22°
Solution:
57°
Use right triangle trigonometry to find x. (Law of sines could also be used.)
sin 57° =
𝑥=
x
21.3
𝑥
21.3cm

21.3
sin 57°
13.4
cm
𝑥 = 25.4
Use Law of Sines to find 𝜃.
𝑥
sin 𝑋
=
𝑦
sin 𝑌
=
𝑧
sin 𝑍
or
sin 𝑋
𝑥
=
sin 𝑌
𝑦
=
sin 𝑍
𝑧
25.4
13.4
=
sin 𝜃 sin 22°
13.4 sin 𝜃 = 25.4 sin 22°
sin 𝜃 =
25.4 sin 22°
13.4
sin 𝜃 = .7101
𝜃 = 𝑠𝑖𝑛−1 . 7101
𝜃 = 45°
However, 𝜃 is obtuse so
𝜽 = 𝟏𝟖𝟎° − 𝟒𝟓° = 𝟏𝟑𝟓°
2009
1.
A triangular sign needs to be painted. The sides measure 60cm, 70cm, and 100cm.
a) Determine the value of  and the area of the triangle.
Solution:
Use Law of Cosines to find 𝜃.
2
2
2
𝑎 = 𝑏 + 𝑐 − 2𝑏𝑐 ∙ cos 𝜃
1002 = 702 + 602 − 2 70 60 cos 𝜃
10000 = 4900 + 3600 − 8400 cos 𝜃
10000 = 8500 − 8400 cos 𝜃
10000 − 8500 = −8400 cos 𝜃
1500 = −8400𝑐𝑜𝑠𝜃
1500
= cos 𝜃
−8400
−.1786 = cos 𝜃
𝜃 = 𝑐𝑜𝑠 −1 −.1786
𝜃 = 100°
70cm

60cm
100cm
b) If a can of paint costs $7.50 and a can of paint cover 520cm², determine the cost to paint the sign assuming you must
purchase entire cans (no partial cans).
Solution:
1
Use the are formula: 𝐴𝑟𝑒𝑎 = 2 𝑎𝑏 sin 𝐶
1
70 60 sin 100°
2
2068 .1
= 2068.1 𝑐𝑚2
Cans of paint = 520 = 3.977 = 4
𝐴𝑟𝑒𝑎 =
Cost = 4 × 7.50 = $30.00
Try:
1. From two drilling platforms A and B, 50 km apart on the Grand Banks, a fishing vessel, S, is sighted. If SAB  51
and SBA  35 , find the distance from the vessel to the nearest platform.
m
B
5
0
k
35°
A
51°
S
2. A soccer net is 8 m wide. If a player is 15 m from the closer post and 21 m from the farthest post, what is the angle, x,
in which he must kick the ball in order to score?
8m
21 m
15 m
x