Linear Functions and
Systems of Equations Unit
(Level IV Graduate Math)
Draft
(NSSAL)
C. David Pilmer
©2009
(Last Updated: Dec 2011)
This resource is the intellectual property of the Adult Education Division of the Nova Scotia
Department of Labour and Advanced Education.
The following are permitted to use and reproduce this resource for classroom purposes.
• Nova Scotia instructors delivering the Nova Scotia Adult Learning Program
• Canadian public school teachers delivering public school curriculum
• Canadian nonprofit tuition-free adult basic education programs
The following are not permitted to use or reproduce this resource without the written
authorization of the Adult Education Division of the Nova Scotia Department of Labour and
Advanced Education.
• Upgrading programs at post-secondary institutions
• Core programs at post-secondary institutions
• Public or private schools outside of Canada
• Basic adult education programs outside of Canada
Individuals, not including teachers or instructors, are permitted to use this resource for their own
learning. They are not permitted to make multiple copies of the resource for distribution. Nor
are they permitted to use this resource under the direction of a teacher or instructor at a learning
institution.
Acknowledgements
The Adult Education Division would also like to thank Eric Tetford (Burridge Campus) for
piloting this resource and offering suggestions during its development.
Table of Contents
Introduction…………………………………………………………………………...
Negotiated Completion Date………………………………………………………….
The Big Picture……………………………………………………………………….
Course Timelines……………………………………………………………………..
ii
iv
v
vi
Introduction to Linear Functions ……………………………………………………..
Find the Equation Given Points and/or Intercepts ……………………………………
Applications of Linear Functions …………………………………………………….
Parallel and Perpendicular Lines ……………………………………………………..
Solving Systems of Equations Graphically …………………………………………..
Solving Systems by Substitution ……………………………………………………..
Solving Systems Using Elimination ………………………………………………….
Inconsistent and Equivalent Equations ……………………………………………….
Break Even Points ……………………………………………………………………
Putting It Together ……………………………………………………………………
1
9
20
28
30
36
44
55
58
61
Post-Unit Reflection ………………………………………………………………….
Additional Practice: Explicit Form (Optional) ……………………………………….
Answers ………………………………………………………………………………
68
69
70
NSSAL
©2009
i
Draft
C. D. Pilmer
Introduction
The Linear Functions and Systems of Equations Unit is one of the four optional units adult
learners may choose from in Level IV Graduate Math. The material in this unit builds on
concepts covered in the Graphs and Functions Unit and relies more heavily on algebraic
procedures. Learners are asked on a few occasions to initially complete questions very similar to
those found in the Graphs and Functions Unit. This is done to review previously learned
concepts and to bridge the gap to the new material.
Negotiated Completion Date
After working for a few days on this unit, sit down with your instructor and negotiate a
completion date for this unit.
Start Date:
_________________
Completion Date:
_________________
Instructor Signature: __________________________
Student Signature:
NSSAL
©2009
__________________________
ii
Draft
C. D. Pilmer
The Big Picture
The following flow chart shows the five required units and the four optional units (choose two of
the four) in Level IV Graduate Math. These have been presented in a suggested order.
Math in the Real World Unit (Required)
• Fractions, decimals, percents, ratios, proportions, and
signed numbers in real world applications
• Career Exploration and Math
Solving Equations Unit (Required)
• Solve and check equations of the form Ax + B = Cx + D ,
A = Bx 2 + C , and A = Bx 3 + C .
Consumer Finance Unit (Required)
• Simple Interest and Compound Interest
• TVM Solver (Loans and Investments)
• Credit and Credit Scores
Graphs and Functions Unit (Required)
• Understanding Graphs
• Linear Functions and Line of Best Fit
Measurement Unit (Required)
• Imperial and Metric Measures
• Precision and Accuracy
• Perimeter, Area and Volume
Choose two of the four.
Linear
Functions and
Linear Systems
Unit
Trigonometry
Unit
Statistics Unit
ALP Approved
Projects
(Complete 2 of
the 5 projects.)
Note:
You are not permitted to complete four ALP Approved Projects and thus avoid selecting from
the Linear Functions and Linear Systems Unit, Trigonometry Unit, or Statistics Unit.
NSSAL
©2009
iii
Draft
C. D. Pilmer
Course Timelines
Graduate Level IV Math is a two credit course within the Adult Learning Program. As a two
credit course, learners are expected to complete 200 hours of course material. Since most ALP
math classes meet for 6 hours each week, the course should be completed within 35 weeks. The
curriculum developers have worked diligently to ensure that the course can be completed within
this time span. Below you will find a chart containing the unit names and suggested completion
times. The hours listed are classroom hours.
Unit Name
Minimum
Completion Time
in Hours
24
20
18
28
24
20
20
Total: 154 hours
Math in the Real World Unit
Solving Equations Unit
Consumer Finance Unit
Graphs and Functions Unit
Measurement Unit
Selected Unit #1
Selected Unit #2
Maximum
Completion Time
in Hours
36
28
24
34
30
24
24
Total: 200 hours
As one can see, this course covers numerous topics and for this reason may seem daunting. You
can complete this course in a timely manner if you manage your time wisely, remain focused,
and seek assistance from your instructor when needed.
NSSAL
©2009
iv
Draft
C. D. Pilmer
Introduction to Linear Functions
As you learned in a previous unit linear functions form straight lines but do not include
horizontal or vertical straight lines.
All of the following are linear functions.
y=
y = 2x + 3
3
x−2
4
1
x
6
y = −1.48 x + 3
y=−
y = x2 + 3
y = 3 sin 120 x°
None of the following are linear functions.
y=3
x=5
All linear functions can be written in the form y = mx + b where m is the slope (i.e. steepness of
the line) and b is the y-intercept (i.e. the location where the line crosses the y-axis). The slope,
m, of a linear function cannot be equal to zero.
Linear Functions:
y = 3 x + 5 where the slope is 3 and the y-intercept is 5
y = −5 x + 9 where the slope is -5 and the y-intercept is 9
y = x − 4 where the slope is 1 and the y-intercept is -4
y=−
2
2
x where the slope is − and the y-intercept is 0
3
3
(Variables other than x and y are often used in many real world applications.)
d = 0.2t + 6 where the slope is 0.2 and the d-intercept is 6
h = −1.9t + 20 where the slope is -1.9 and the h-intercept is 20
v = 12.56h where the slope is 12.56 and the v-intercept is 0
NSSAL
©2009
1
Draft
C. D. Pilmer
Nonlinear Functions:
y=5
y=
2
−1
x
y = 3x 2 − 7
y=
1
0.4 x + 3
y = 4x + 2
y = sin x
Example 1:
Determine whether each of the following is a linear function.
(a) 3x + 5 y − 10 = 0
(b) 6 x − 4 y = 3(2 x + 1)
(c) 0.7 x − y + 5 = 9
(d) 5( y − 3 x ) = 2 x 2
2
1
(e) 6 xy + 4 = 11x − 1
(f) y + x − 9 = 0
3
4
Answers:
Rearrange each equation to its explicit form (i.e. "y =" form).
(a) 3 x + 5 y − 10 = 0
(b) 6 x − 4 y = 3(2 x + 1)
5 y = −3 x + 10
6x − 4 y = 6x + 3
5 y − 3 x 10
− 4y = 3
=
+
5
5
5
3
Not Linear
y=−
3
4
Linear
y=− x+2
5
(c)
(e)
NSSAL
©2009
0.7 x − y + 5 = 9
− y = −0.7 x + 9 − 5
− y = −0.7 x + 4
− y − 0.7 x 4
=
+
−1
−1
−1
Linear
y = 0.7 x − 4
6 xy + 4 = 11x − 1
6 xy = 11x − 1 − 4
6 xy = 11x − 5
6 xy 11x 5
=
−
6x
6x 6x
11 5
y= −
6 6x
(d)
5( y − 3 x ) = 2 x 2
5 y − 15 x = 2 x 2
5 y = 2 x 2 + 15 x
5 y 2 x 2 15 x
=
+
5
5
5
2
y = x 2 + 3x
5
(f)
Not Linear
2
Not Linear
2
1
y + x = −9
3
4
2 
1 
12 y  + 12 x  = 12(− 9 )
3 
4 
8 y + 3x = −108
8 y = −3x − 108
8 y − 3x 108
=
−
8
8
8
3
27
Linear
y =− x−
8
2
Draft
C. D. Pilmer
When a linear function is written as y = mx + b (i.e. slope y-intercept form), you can quickly
visualize what the straight line graph looks like.
Example 2:
Match the linear function with the
appropriate straight line graph. Do not
generate a table of values or use graphing
technology.
(a) y = 3 x + 2
(b) y = −2 x + 4
(c) y = − x − 3
(d) y = 1.5 x
Line A
Line B
6
5
4
3
2
1
0
-6
-5 -4
-3
-2 -1-1 0
1
2
3
4
5
6
-2
-3
-4
-5
-6
Line D
Line C
Answers:
• The linear function y = 3 x + 2 has a slope of 3 and a y-intercept of 2. Since the yintercept is 2, the straight line intersects the y-axis at 2. Since the slope is a positive
number, then we know that the graph is increasing (as we move from left to right along
the line, the line goes upwards). This equation matches Line A.
•
The linear function y = −2 x + 4 has a slope of -2 and a y-intercept of 4. Since the yintercept is 4, the straight line intersects the y-axis at 4. Since the slope is a negative
number, then we know that the graph is decreasing (as we move from left to right along
the line, the line goes downwards). This equation matches Line C.
•
The linear function y = − x − 3 has a slope of -1 and a y-intercept of -3. Since the yintercept is -3, the straight line intersects the y-axis at -3. Since the slope is a negative
number, then we know that the graph is decreasing (as we move from left to right along
the line, the line goes downwards). This equation matches Line D.
•
The linear function y = 1.5 x has a slope of 1.5 and a y-intercept of 0. Since the yintercept is 0, the straight line intersects the y-axis at 0 (i.e. passes through the origin).
Since the slope is a positive number, then we know that the graph is increasing (as we
move from left to right along the line, the line goes upwards). This equation matches
Line B.
NSSAL
©2009
3
Draft
C. D. Pilmer
Example 3:
Graph 4 x − 3 y − 6 = 0 without generating a table of values.
Answer:
• Change to the slope y-intercept form.
4x − 3y − 6 = 0
− 3 y = −4 x + 6
4
y = x−2
3
• Plot the y-intercept (-2) on the graph.
• Starting at the y-intercept, rise and run the
appropriate amount based on the slope.
4
Because the slope is , you will run 3
3
(over to the right 3) and rise 4 (go up 4).
Plot this new point.
• Draw a straight line that passes through the
two points.
y
5
4
3
2
y-intercept of -2
1
rise = 4
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
-1
-2
-3
run = 3
-4
-5
x
Questions
1. Which of these graphs represent linear functions?
(a)
(b)
(c)
(e)
(f)
(g)
(d)
(h)
2. Convert the equations of each of these linear functions to their slope y-intercept forms. State
the slope and the y-intercept.
(a) 5 x + 15 y − 30 = 0
(b) 9 x − 2 y + 14 = 0
NSSAL
©2009
4
Draft
C. D. Pilmer
(c) 7 x − 3 y − 5 = x + 4
(d) 5 x + 2 y + 6 = 8 x − 8
(e) 2(3 x + 5 y ) = 9 x − 20
(f) 3(2 x − 5 y ) = x + 30
(g) 2( y + 4 x ) = 3( x + 4 )
(h) 3(2 y − 5 x ) = 4( x − 6 )
(i) 0.7 x + 0.6 y − 48 = 0
(j) x − 0.3 y + 1 = 3 x + 1
(k)
NSSAL
©2009
2
1
y + x − 29 = 1
5
3
(l)
5
5
1
x + y = 24
4
6
Draft
C. D. Pilmer
3. Which of these equations represent a linear function?
2
(a) y = x + 17
(b) y = − x + 6
9
(d) y =
2
+ 17
9x
(g) y = (3 x + 1)
2
(j) y = 7 x −
2
5
(e) y = 11 − 5 x
(h) y =
13
x
5
(k) y = −
(c) y = 4 x 2 − 7
(f) y = −0.579 x + 0.46
(i) y = −5 x +
2
9
2
x
(l) y = 7 + 0.4 x
4. Rearrange each equation into its explicit form (if possible). Identify those that are linear
functions.
(a) 4 x + 2 y − 17 = 0
(b) 5 x − 3 y − 6 = 0
(c) 3( x + 2 y ) = 12
(d) 14 x − 7 y + 50 = 1
(e) 2(3 y + 2 x ) = 24 + 4 x
(f)
NSSAL
©2009
6
2
1
y− x=6
3
2
Draft
C. D. Pilmer
(g) 0.3 x − 1.4 + 0.7 y = 0
(h) 4 x − 6 y + 2 = 2( x − 3 y )
(i) 7 − 2 xy + 8 x = 0
(j) 14 x 2 + 6 x + 2 y − 4 = 0
(k) 0.3(4 + 2 y ) − 5 x = 7
(l) x 3 −
1
y + 1 = 2x
6
5. Match the equation with the appropriate straight line graph.
Equation
1
y =− x+4
3
Line
6
A
B
5
4
y = −2 x − 3
3
C
2
1
x−3
2
4
y = x+4
3
y=
1
0
-6 -5 -4 -3 -2 -1-1 0
1
2
3
4
5
6
-2
-3
-4
y = 3x
-5
-6
NSSAL
©2009
D
7
E
Draft
C. D. Pilmer
6. Graph each of these linear functions on the coordinate system that has been provided. Do not
generate a table of values.
(a)
5x + 3 y + 6 = 4 x
(b)
3x + y − 3 = 1
7
7
6
6
5
5
4
4
3
3
2
2
1
1
0
-7
-6
-5
-4
-3
-2
-1
-1
0
0
1
2
3
4
5
6
7
-7
-6
-5
-4
-3
-2
-1
-1
-2
-2
-3
-3
-4
-4
-5
-5
-6
-6
-7
-7
0
1
2
3
4
5
6
7
7. Graph each of these tables of values without generating a table. Look at the coordinate
systems that have been provided for each question. Notice that the scales on the two axes are
different. These scales were chosen to make the graphing of the functions easier.
(a)
7 x − 50 y − 300 = 0
(b)
40 x + 2 y − 60 = 0
7
50
6
40
5
4
30
3
20
2
10
1
0
-100
-75
-50
-25
-1
0
0
25
50
75
100
-4
-3
-2
-1
0
1
2
3
4
-10
-2
-3
-20
-4
-30
-5
-40
-6
-7
-50
If you feel that you need additional practice changing equations into their explicit form, you can
complete the optional questions titled Additional Practice: Explicit Form found in the appendix
of this unit.
NSSAL
©2009
8
Draft
C. D. Pilmer
Find the Equation Given Points and/or Intercepts
Up to this point, we have found the equation of a linear function by:
(1) looking for patterns in the table of values.
Example A
Determine the equation of
the linear function given
the table of values on the
right.
x
-4
-2
0
2
y
1
4
7
10
Answer:
diff.
2
2
2
2
x
-4
-2
0
2
4
y
1
4
7
10
13
diff.
3
3
3
3
As the x-values change by increments of 2, there is a
common difference of 3 between successive y-values. It’s a
linear function.
The y-intercept occurs when x = 0, so the y-intercept is at 7.
The difference between the successive x-values ( ∆x ) and
difference between successive y-values ( ∆y ) gives you the
3
∆y
slope. The slope equals
so it equals
in this case.
∆x
2
The equation is y =
3
x +7.
2
(2) finding critical information contained in the graph of the function.
Example B
Determine the equation of the function based on the graph of the function.
Answer:
The y-intercept is 7. You now can write the
equation as y = mx + 7 . You still need to find
the slope, m.
11
y10
y-intercept = 7
9
8
run = 1
7
6
rise = -2
5
4
3
2
1
0
-2
-1
-1
0
1
2
3
4
The x-values are changing by 1. That means we
have a run of 1. The y-values are changing by -2.
That means we have a rise of -2.
rise
slope =
run
−2
=
1
x
= −2
Equation: y = −2 x + 7
NSSAL
©2009
9
Draft
C. D. Pilmer
In this section, we are going to concentrate on techniques that require algebraic manipulation.
We will still be working with the slope y-intercept form (i.e. y = mx + b ) of linear functions;
however, the slope will often be found using a new formula and the y-intercept will often be
found using algebra.
When we learned how to determine the equation of a linear function given a table of values, we
learned that the slope of a linear function can be found using the following.
slope =
difference between successive y - values
difference between successive x - values
or
slope =
change in y - values
change in x - values
This could be restated as the following.
slope =
∆y
∆x
- where ∆ ("delta") means the "change in"
We are now going to expand upon this last equation for the slope. We can now state the
following.
slope =
y 2 − y1
x 2 − x1
- this is the new formula that we will be using in this section
Example 1
Determine the slope of the linear function passing through the points (-5, 6) and (2, -9).
Answer:
We can now solve this question two ways; using the new formula or the method involving a
table of values. Normally we need more than two points to use the table of values method
because we need to see the common differences to insure that we're working with a linear
function. We do not have to worry about this because the question said explicitly that we are
dealing with a linear function.
New Method
Old Method
(-5, 6) and (2, -9)
(x1 , y1 )
(x2 , y 2 )
y − y1
slope = 2
x 2 − x1
−9−6
=
2 − (− 5)
15
− 15
or −
=
7
7
diff.
7
x
-5
2
y
6
-9
diff.
-15
∆y
∆x
− 15
15
or −
=
7
7
slope =
As you move into higher level mathematics courses, you will see why it is important to know
this new method (i.e. the new slope formula).
NSSAL
©2009
10
Draft
C. D. Pilmer
Example 2
Determine the equation of the line that passes through the points (1, 7) and (-3, -5).
Answer:
We need to find the slope, m, and the y-intercept, b, in the equation y = mx + b .
First find the slope using the new formula from the previous page.
(1, 7) and (-3, -5)
(x1 , y1 )
(x2 , y 2 )
y − y1
slope = 2
x 2 − x1
−5−7
=
− 3 −1
− 12
=
−4
=3
Put the numerical value for the slope into the equation. Now substitute the x value and y
value for one of the points into the equation so that we can solve for b, the y-intercept.
y = 3x + b
← (1, 7)
7 = 3(1) + b
7 = 3+b
4=b
Now that we have the slope and y-intercept, we can write the equation.
y = 3x + 4
Example 3
Determine the equation of the line that passes through the points (6, -3) and (-9, 7).
Answer:
(6, -3) and (-9, 7)
(x1 , y1 )
(x2 , y 2 )
y − y1
slope = 2
x 2 − x1
7 − (− 3)
=
−9−6
10
=
− 15
2
=−
3
NSSAL
©2009
2
y = − x+b
← (6, -3)
3
2
− 3 = − (6 ) + b
3
− 3 = −4 + b
1= b
2
y = − x +1
3
11
Draft
C. D. Pilmer
The other thing we must talk about, before looking at any more worked examples, is y-intercepts
and x-intercepts.
The y-intercept is the
y-coordinate of the point where
the line crosses the y-axis.
Consider Lines A and B drawn
on the coordinate system. The
y-intercept of line A is 3. When
written as an ordered pair, the
point is (0, 3). The y-intercept
of line B is 6. When written as
an ordered pair, the point is
(0, 6).
The x-intercept is the
x-coordinate of the point where
the line crosses the x-axis.
Consider Lines A and B drawn
on the coordinate system. The
x-intercept of line A is 2. When
written as an ordered pair, the
point is (2, 0). The x-intercept
of line B is -3. When written as
an ordered pair, the point is
(-3, 0).
Line A
Line B
7
6
5
4
3
x-intercept
2
1
0
-7 -6 -5 -4 -3 -2 -1-1 0 1
-2
-3
-4
-5
-6
-7
y-intercept
y-intercept
x-intercept
2 3
4 5
6 7
Two Important Facts
• The y-intercept occurs when the x-coordinate equals zero. Therefore if we know a
function has a y-intercept of 4, then we can say that the function passes through the point
(0, 4).
• The x-intercept occurs when the y-coordinate equals zero. Therefore if we know a
function has a x-intercept of -7, then we can say that the function passes through the point
(-7, 0).
NSSAL
©2009
12
Draft
C. D. Pilmer
Example 4
Determine the x-intercept and y-intercept of the function 3 x − 7 y + 42 = 0 .
Answers:
The x-intercept occurs when the ycoordinate equals zero. Substitute 0 in for
y, and solve for x.
The y-intercept occurs when the xcoordinate equals zero. Substitute 0 in for
x, and solve for y.
3 x − 7 y + 42 = 0
3(0) − 7 y + 42 = 0
− 7 y + 42 = 0
3 x − 7 y + 42 = 0
3 x − 7(0) + 42 = 0
3 x + 42 = 0
3 x = −42
− 7 y = −42
y=6
x = −14
The x-intercept is -14.
The y-intercept is 6.
Example 5
Determine the equation of the linear function that passes through the point (-10, 9) and has a
x-intercept of -25.
Answer:
Since we have a x-intercept of -25, we know that the line passes through (-25, 0). This
question can now be solved using the procedure from Examples 2 and 3 where we were
supplied with the coordinates of two points on the linear function.
(-10, 9) and (-25, 0)
(x1 , y1 )
(x2 , y 2 )
y − y1
slope = 2
x 2 − x1
0−9
=
− 25 − (− 10 )
−9
=
− 15
3
=
5
NSSAL
©2009
3
x+b
← (-25, 0)
5
3
0 = (− 25) + b
5
0 = −15 + b
15 = b
y=
y=
3
x + 15
5
13
Draft
C. D. Pilmer
Example 6
Determine the equation of the linear function that has a y-intercept of 12 and a x-intercept of 18.
Answer:
y-intercept of 12 means that we are dealing with the point (0, 12)
x-intercept of 18 means that we are dealing with the point (18, 0)
(0, 12) and (18, 0)
(x1 , y1 )
(x2 , y 2 )
y − y1
slope = 2
x 2 − x1
0 − 12
=
18 − 0
− 12
=
18
2
=−
3
We do not have to calculate b, the
y-intercept, because it was supplied in the
question.
2
y = − x+b
3
2
y = − x + 12
3
Example 7
Determine the equation of the linear function that has a slope of -3 and a x-intercept of 2.
Answer:
We do not need to calculate the
slope, m, because it was supplied in
the question.
y = −3 x + b
0 = −3(2) + b
0 = −6 + b
6=b
← (2, 0)
y = −3 x + 6
Example 8
Determine the x-intercept of the linear function that passes through (-6, -20) and (3, -8).
Answer:
We need to first find the equation of the function and later find the x-intercept.
4
4
y − y1
y = x+b
y = x − 12
← (-6, -20)
slope = 2
3
3
x 2 − x1
4
4
− 8 − (− 20 )
− 20 = (− 6) + b
0 = x − 12
=
3
3
3 − (− 6 )
− 20 = −8 + b
4
12
12 = x
−12 = b
=
3
9
4 
4
4
3(12) = 3 x 
=
y = x − 12
3 
3
3
36 = 4 x
x = 9 ← x - intercept
NSSAL
©2009
14
Draft
C. D. Pilmer
Questions:
1. Determine the equation of the linear function that passes through (1, 6) and (-3, 14).
2. Find the equation of the straight line that passes through the points (2, -4) and (12, 1).
3. If a linear function passes through the points (6, -11) and (-9, 14), determine the equation of
that function.
4. If the x-intercept of a linear function is 32 and the y-intercept of the same function is -24,
determine the equation of the function.
5. Determine the equation of the linear function that has y-intercept of 4 and passes through the
point (7, 2).
NSSAL
©2009
15
Draft
C. D. Pilmer
6. Determine the equation of the linear function that has a slope of -5 and passes through the
point (-2, 18)
7. Find the equation of the straight line graph that passes through (6, -20) and has a x-intercept
of -9.
8. A linear function passes through the points (-4, 1) and (6, 16). Determine its equation.
9. The slope of a linear function is
7
. It's x-intercept is 10. Determine the equation of the
5
function.
10. Find the equation of the straight line graph that passes through (16, -4) and has a y-intercept
of 6.
NSSAL
©2009
16
Draft
C. D. Pilmer
11. Find the equation of the straight line that passes through the points ( 7, -7) and (-14, 20).
12. Determine the equation of the linear function that has a x-intercept of 18 and passes through
the point (-9, -24).
13. Determine the equation of the linear function that has a x-intercept of -21 and a y-intercept of
-9.
14. Determine the x-intercept and y-intercept of each of these linear functions.
(a) 2 x + 5 y − 30 = 0
(b) 4 x − 3 y − 24 = 0
NSSAL
©2009
17
Draft
C. D. Pilmer
(c) 28 − 2 y + 7 x = 0
(d) 6 y − 5 x − 48 = 12
15. Determine the equation of the straight line that passes through the points (8, 24), (-12, -21)
and (-4, -3).
16. Determine the equation of the linear function that has a x-intercept of 4, a y-intercept of 14
and passes through the point (10, -21).
17. What can you say about questions 15 and 16?
NSSAL
©2009
18
Draft
C. D. Pilmer
18. Determine the x-intercept of the linear function that passes through (30, 5) and (-18, -35).
19. Determine the x-intercept of the linear function that passes through (12, -14) and (3, 7).
NSSAL
©2009
19
Draft
C. D. Pilmer
Applications of Linear Functions
We have learned how to determine the equations of linear functions in a variety of ways.
• Given a table of values (We did this in previous unit)
• Given a graph (We did this in a previous unit)
• Given points and/or intercepts (We did this in this unit)
In this section, you will put the last set of skills to use to solve a variety of application questions.
Example 1:
Water is being pumped out of your neighbour's cylindrical above ground pool at a constant rate.
At t = 5 minutes, the water is 44 inches deep. At t = 25 minutes, the water is 28 inches deep.
Assume that the depth of the water and the time are related by a linear function.
(a) Determine the equation of the function that describes the depth, d, of the water in terms of
time, t.
(b) What is the depth of the water at t = 18 minutes?
(c) When will the water be 22 inches deep?
(d) When will the pool be empty?
(e) At what rate is the depth of the water changing?
(f) What was the initial depth of the water in the pool?
(g) If the water was initially 46 inches deep and being drained at a rate of 1.5 inches per hour,
what would be the equation describing the depth in terms of time?
Answers:
(a) When one writes y = mx + b , we say that the dependent variable y is expressed in terms
of the independent variable x. Our question states that we need to find the equation that
describes the depth, d, of the water in terms of time, t. That means the depth is the
dependent variable and time is the independent variable.
(t, d)
(5, 44)
(25, 28)
∆d
∆t
28 − 44
m=
25 − 5
− 16
m=
20
4
m=−
5
m=
d = mt + b
4
d = − t + b ← (5,44 )
5
4
44 = − (5) + b
5
44 = −4 + b
48 = b
4
Therefore: d = − t + 48
5
NSSAL
©2009
20
Draft
C. D. Pilmer
4
(b) d = − t + 48
5
4
d = − (18) + 48
5
d = −14.4 + 48
d = 33.6
The depth of the water at t = 18 minutes is 33.6 inches.
4
(c) d = − t + 48
5
4
22 = − t + 48
5
4
22 − 48 = − t
5
4
− 26 = − t
5
 4 
5(− 26) = 5 − t 
 5 
− 130 = −4t
The depth of the water will be 26 inches when t equals 32.5 minutes.
t = 32.5
(d) The pool will be empty when the depth, d, is equal to zero.
4
d = − t + 48
5
4
0 = − t + 48
5
4
− 48 = − t
5
 4 
5(− 48) = 5 − t 
 5 
− 240 = −4t
The pool will be empty when t equals 60 minutes.
t = 60
(e) The rate of change for any linear function is equivalent to the slope. Since our slope is
4
− , we can say that the pool depth is dropping by 4 inches every 5 minutes.
5
(f) The initial depth is equivalent to the depth intercept. That means that the depth of the
water was initially 48 inches.
(g) d = −1.5t + 46
NSSAL
©2009
21
Draft
C. D. Pilmer
Example 2:
Ryan is renting a car for the day. The rental company charges a flat rate plus an amount per
kilometre driven. If he drives 150 km, he must pay $54.99. If he drives 250 km, he must pay
$77.49. Assume that the cost of the rental car varies linearly with time.
(a) Determine the equation that describes the cost, c, of the rental in terms of the distance, d,
driven.
(b) What does the c-intercept represent in this situation?
(c) What does the slope represent in this situation?
(d) If he only has $100 to spend on the rental car, how far can he travel?
(e) How much does it cost to rent the car for a round trip that covers 620 km?
(f) If the rental cost was calculated using a flat rate was $27.99 plus a charge of $0.19 per
kilometre, what would be the equation describing the rental cost in terms of distance driven?
Answers:
(a) Our question states that we need to find the equation that describes the cost, c, in terms of
distance, d. That means the cost is the dependent variable and distance is the
independent variable.
(d, c)
c = md + b
∆c
m
=
(150, 54.49)
∆d
c = 0.23d + b ← (150, 54.49 )
(250, 77.49)
77.49 − 54.49
54.49 = 0.23(150 ) + b
m=
250 − 150
54.49 = 34.5 + b
23
19.99 = b
m=
100
Therefore:
c = 0.23d + 19.99
m = 0.23
(b) The c-intercept represents the flat rate ($19.99) charged by the rental company.
(c) The slope represents the cost per kilometre driven ($0.23 per km).
(d) c = 0.23d + 19.99
100 = 0.23d + 19.99
80.01 = 0.23d
80.01 0.23d
=
0.23
0.23
He can travel 347 kilometres.
d = 347.9
(e) c = 0.23d + 19.99
c = 0.23(620 ) + 19.99
c = 142.6 + 19.99
The rental car will cost $162.59.
c = 162.59
(f) c = 0.19d + 27.99
NSSAL
©2009
22
Draft
C. D. Pilmer
Questions:
1. A patient is hooked up to an intravenous (I.V.) bag. It gradually dispenses medication to the
patient. Two hours after the I.V. bag is hooked up, 840 mL of fluid remains in the bag. Nine
hours after the bag is hooked up, 280 mL remains.
(a) What is the equation of the linear function that describes the amount, a, of fluid
remaining in the I.V. bag in terms of time, t?
(b) How much fluid is in the bag five hours after it was hooked up?
(c) How much fluid was initially in the bag?
(d) At what time will there be 160 mL of fluid in the bag?
(e) When will the bag be empty?
(f) What is the numerical value of the slope and what does it represent in this situation?
(g) If the I.V. bag initially held 500 mL of fluid and drained at a rate of 60 mL per hour,
what would be the equation that described the amount in terms of time?
NSSAL
©2009
23
Draft
C. D. Pilmer
2. A cyclist is traveling on level ground at a constant speed. She passes Markus as he is
standing on the road side. Six seconds after she has passed Markus, she is 80 m from him.
She is 200 m from him fifteen seconds after she passed him.
(a) Assuming that the distance between the cyclist and Markus varies linearly with time,
determine the equation of the function that describes the distance, d, in terms of time, t.
(b) What is the distance between the cyclist and Markus at t = 20 seconds?
(c) Could the linear function you determined in (a) be used to predict the distance between
the cyclist and Markus at t = 600 seconds (10 minutes)? Why or why not?
(d) When will the distance between the cyclist and Markus be 360 m?
(e) Is the slope positive or negative? Why does this make sense for this situation?
(f) If the cyclist was traveling 0.5 metres per second faster, what would be the equation
describing the distance in terms of time?
NSSAL
©2009
24
Draft
C. D. Pilmer
3. Vivian receives a flat salary for her 40 hour work week. If she works overtime, she receives
double time (i.e. paid an hourly rate that is double the normal hourly rate). If she works a full
week plus three hours of overtime, she receives $644 (before deductions). When she works a
full week plus seven hours of overtime, she receives $756 (before deductions).
(a) Determine the equation of the straight line that describes Vivian's weekly earnings, e, in
terms of the number of overtime hours, h, she works.
(b) What is Vivian's flat salary for a regular work week?
(c) If Vivian earned $868 (before deductions) in one week, how many overtime hours did
she work?
(d) What does she receive per hour for overtime work?
(e) Is the slope positive or negative? Why does this make sense for this situation?
(f) If the equation was e = 18h + 360 , how would the situation have changed?
NSSAL
©2009
25
Draft
C. D. Pilmer
4. Donna owns a pellet stove. That morning she partially filled the hopper on the stove with
pellets. They gradually burned off during the day. Two hours after she partially filled the
hopper, 4.4 kg of pellets remained in the hopper. Five hours after she partially filled the
hopper, 2.3 kg of pellets remained in the hopper. Assume that the mass of pellets remaining
in the hopper varies linearly with time.
(a) Determine the equation of the function that describes the mass, M, of pellets in the
hopper in terms of time, t.
(b) At what rate was the stove burning pellets?
(c) How many kilograms of pellets remained in the hopper three hours after the hopper was
partially filled?
(d) How many kilograms of pellets were initially in the hopper?
(e) When will there be 0.9 kg of pellets in the hopper?
(f) When will the hopper be empty?
(g) If the hopper could hold 0.5 kg more of pellets and burned the pellets at a rate 0.6 kg per
hour, what would be the new equation of the linear function?
NSSAL
©2009
26
Draft
C. D. Pilmer
5. Margaret's cousin from the United States is visiting Nova Scotia. The cousin is not familiar
with the metric system and gets confused when weather reports state the temperature in
degrees Celsius. Margaret wants to help her cousin by creating an equation that converts
Celsius to Fahrenheit. Margaret knows that 0oC is equivalent to 32oF, and 100oC is
equivalent to 212oF.
(a) Determine the equation of the straight line that describes degrees Fahrenheit, F, in terms
of degrees Celsius, C.
(b) Use your equation to convert 34oC to degrees Fahrenheit.
(c) Use your equation to convert 65oF to degrees Celsius.
NSSAL
©2009
27
Draft
C. D. Pilmer
Parallel and Perpendicular Lines
In each case we have been provided with two linear functions. Graph both functions on the same
coordinate system.
(a) y =
3
3
x + 2 and y = x − 4
5
5
(b) y = −
4
4
x + 6 and y = − x
3
3
7
7
6
6
5
5
4
4
3
3
2
2
1
1
0
-7
-6
(c) y =
-5
-4
-3
-2
-1
-1
0
0
1
2
3
4
5
6
7
-7
-6
-5
-4
-3
-2
-1
-2
-2
-3
-3
-4
-4
-5
-5
-6
-6
-7
-7
2
3
x − 5 and y = − x + 1
3
2
(d) y = −
-6
NSSAL
©2009
-5
-4
-3
-2
-1
7
7
6
6
5
5
4
4
3
3
2
2
1
1
-1
0
1
2
3
4
5
6
7
3
4
5
6
7
1
2
x and y = x + 3
2
1
0
-7
-1
0
0
1
2
3
4
5
6
7
-7
-6
-5
-4
-3
-2
-1
-1
-2
-2
-3
-3
-4
-4
-5
-5
-6
-6
-7
-7
28
0
1
2
Draft
C. D. Pilmer
Questions:
1. Describe the orientation of the lines in parts (a) and (b) on the previous page.
2. Examine the equations in parts (a) and (b). How are the equations similar? What conclusion
we make?
3. Given the linear function y =
4
x + 6 , supply the equations of three linear functions that are
7
parallel to it.
4. Describe the orientation of the lines in parts (c) and (d) on the previous page.
5. Examine the equations in parts (c) and (d). How are the equations related? What conclusion
we make?
6. Given the linear function y =
2
x − 4 , supply the equations of three linear functions that are
5
perpendicular to it.
8
7. Given the linear function y = − x + 2 , supply the equations of three linear functions that
3
are perpendicular to it.
8. Given the linear function y = −3 x + 8 , which of the listed functions below are:
(a) parallel to it?
_________________
(b) perpendicular to it?
_________________
Listed Functions:
1
(i) y = x − 2
3
(ii) y = −
(iv) y = −3 x
(v) y =
NSSAL
©2009
5
x+8
7
1
x + 12
3
29
(iii) y = −3 x + 4
(vi) y = 2 x −
1
8
Draft
C. D. Pilmer
Solving Systems of Equations Graphically
In the previous section, Interpretation Involving Two Linear Functions, you had to find the point
of intersection between two linear functions. Although you didn’t realize it, you were actually
solving a system of two equations in two variables when you found the point of intersection.
The two equations were the equations of the two linear functions. The two variables were the
two coordinates for the point of intersection.
There are many situations when we need to find the point of intersection that can also be called
the point that satisfies both equations simultaneously. This second part of statement is best
explained using an example.
Example 1:
Solve the following system of equations graphically. Check your answer.
5x + 2 y = 8
x− y=3
Answer:
Change both equations to the slope, y-intercept form (i.e. y = mx + b ).
5x + 2 y = 8
x− y=3
2 y = −5 x + 8
− y = −x + 3
2 y − 5x 8
y = x−3
=
+
2
2
2
5
y=− x+4
2
Plot both graphs on the same coordinate system
by first plotting the y-intercepts and then rising
and running from that point based on the slope.
5
y=− x+4
2
y-intercept = 4
4
3
y = x−3
2
1
y-intercept = -3
0
-2
rise − 5
slope =
=
run
2
y
5
rise 1
slope =
=
run 1
-1
0
-1
1
2
3
4
x
(2, -1)
-2
-3
-4
The point of intersection is (2, -1). That means that the solution to this system of
equations is x = 2 and y = -1.
To check the answer, substitute the values of the variables into both of the original
questions and see if they satisfy the equations (i.e. see if both sides of the equation are
still equal to each other).
NSSAL
©2009
30
Draft
C. D. Pilmer
Check:
5x + 2 y = 8
5(2 ) + 2(− 1) = 8
10 + (−2) = 8
8 = 8 satisfies the first equation
x− y=3
2 − (−1) = 3
3 = 3 satisfies the second equation
The answer x = 2 and y = -1 is correct.
Example 2:
Solve the following system of equations graphically. Check your answer.
x − 3 y = −18
2 x + y = −1
Answer:
Change both equations to the slope, y-intercept form (i.e. y = mx + b ).
x − 3 y = −18
2 x + y = −1
− 3 y = − x − 18
y = −2 x − 1
− 3 y − x 18
=
−
−3 −3 −3
1
y = x+6
3
y-intercept = 6
slope =
rise 1
=
run 3
y
y-intercept = -1
slope =
8
rise − 2
=
run
1
7
(-3, 5)
6
5
The point of intersection is (-3, 5). That
means that the solution to this system of
equations is x = -3 and y = 5.
4
3
2
1
0
-5
-4
-3
-2
-1
0
1
2
3
x
-1
-2
-3
Check:
x − 3 y = −18
− 3 − 3(5) = −18
− 3 − 15 = −18
− 18 = −18 satisfies the first equation
2 x + y = −1
2(− 3) + 5 = −1
− 6 + 5 = −1
− 1 = −1 satisfies the second equation
The answer x = -3 and y = 5 is correct.
NSSAL
©2009
31
Draft
C. D. Pilmer
Example 3:
After hurricane Juan, Tom needed to rent a chainsaw to clear up a few small trees that had been
uprooted on his property. Tools-R-Us charges a base fee of $18, plus $6 per hour. Another
company, Tool Depot, charges a base fee of $24, plus $4 per hour.
(a) Determine the equation of the linear function that describes the Tools-R-Us rental charge, C,
in terms of the number of hours, h, that the chainsaw is rented.
(b) Determine the equation of the linear function that describes the Tool Depot rental charge, C,
in terms of the number of hours, h, that the chainsaw is rented.
(c) Determine the point of intersection graphically.
(d) Explain what the point of intersection represents in this situation.
Answers:
(a) C = 6h + 18
(b) C = 4h + 24
C = 6h + 18
C = 4h + 24
y-intercept = 18
y-intercept = 24
slope =
rise 6
=
run 1
slope =
rise 4
=
run 1
The point of intersection is (3, 36).
(d) If Tom rents a chainsaw from either of
these companies for 3 hours, he has to pay
the same amount, $36.
Rental Cost
(c) Graph both linear functions on the same
coordinate system. The most challenging
thing with this question is choosing
appropriate scales on the two axes.
46
44
42
40
38
36
34
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
0
1
2
3
4
5
Time
Questions
1. Determine the slope and y-intercept of each line.
(a) 2 x + 3 y = 15
(b) 8 x − 6 y + 12 = 0
NSSAL
©2009
32
(c) 0.8 x − 0.2 y − 1 = 0
Draft
C. D. Pilmer
2. For each system of equations, determine whether the ordered pair is the solution.
(a) 3 x + 5 y = −1 and 2 x + 7 y = 3 ; (-2, 1)
(b) 4 x + 2 y = 2 and 6 x − y = 12 ; (3, -5)
(c) 5 x − 3 y = 7 and − 2 x + 4 y = −14 ; (-1, -4)
3. Solve each of the following systems graphically and verify your solution.
(a) 2 x + 3 y = 12 and 4 x − 3 y = 6
7
6
5
4
3
2
1
0
-7
-6
-5
-4
-3
-2
-1
-1
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
-2
-3
-4
-5
-6
-7
(b)
x − 2 y = 10 and 5 x + 4 y = 8
7
6
5
4
3
2
1
0
-7
-6
-5
-4
-3
-2
-1
-1
-2
-3
-4
-5
-6
-7
NSSAL
©2009
33
Draft
C. D. Pilmer
(c)
x − y = −4 and 2 x − 5 y = −5
7
6
5
4
3
2
1
0
-7
-6
-5
-4
-3
-2
-1
-1
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
-2
-3
-4
-5
-6
-7
(d)
x − 2 y = −8 and 2 x + y = −1
7
6
5
4
3
2
1
0
-7
-6
-5
-4
-3
-2
-1
-1
-2
-3
-4
-5
-6
-7
(e)
x + y − 1 = 0 and 0.6 x + y + 1 = 0
7
6
(Hint: Express 0.6 as a fraction.)
5
4
3
2
1
0
-7
-6
-5
-4
-3
-2
-1
-1
-2
-3
-4
-5
-6
-7
Scrambled Answer Key for Question 3
(-5, -1)
(3, 2)
(-6, 4)
NSSAL
©2009
(-2, 3)
34
(4, -3)
(5, -4)
Draft
C. D. Pilmer
4. Candice needs to have two small rooms in her house painted. She is going to hire a painter.
The first painter said that he would charge a flat fee of $150 to cover materials and travel,
plus $30 per hour. The second painter said that she would charge a flat fee of $165, plus $25
per hour.
(a) Determine the equation of the linear function that describes the cost, c, of using the first
painter in terms of the number of hours, h, he works.
(b) Determine the equation of the linear function that describes the cost, c, of using the
second painter in terms of the number of hours, h, she works.
(c) Determine the point of intersection graphically.
(d) Explain what the point of intersection represents in this situation.
260
255
250
245
240
235
230
225
220
215
Cost
210
205
200
195
190
185
180
175
170
165
160
155
150
0
1
2
3
4
5
6
7
8
Hours of Labour
NSSAL
©2009
35
Draft
C. D. Pilmer
Solving Systems by Substitution
In the previous section you learned how to solve systems graphically. There are a few problems
with this method.
• It’s a time consuming process.
• Many linear functions are challenging to graph accurately. (e.g. y = 0.165 x + 22.7 )
• If the coordinates of the point of intersection do not appear to be integers, then it is
almost impossible to determine the correct solution using only the graphs.
There are two algebraic methods of solving systems of equation; substitution and elimination. In
this section, we will focus on substitution. With this algebraic technique we substitute one
equation into the other equation such that we change the question from two equations with two
variables to one equation with one variable. This is best understood by looking at a few
examples.
Example 1:
Solve the following system using substitution.
x + 3 y = −2 and 5 x + 4 y = 12
Answer:
x + 3 y = −2
x = −3 y − 2
Choose one of the equations and isolate one
of the variables by expressing that variable in
terms of the other.
In this example, we took the first equation
and isolated x. It was the easiest thing to do.
5 x + 4 y = 12
5(− 3 y − 2) + 4 y = 12
5(− 3 y − 2 ) + 4 y = 12
− 15 y − 10 + 4 y = 12
− 15 y + 4 y = 12 + 10
− 11 y = 22
− 11 y
22
=
− 11 − 11
y = −2
x + 3 y = −2
x + 3(− 2 ) = −2
x − 6 = −2
x = −2 + 6
x=4
NSSAL
©2009
Substitute the expression for x into the other
equation. By doing this, we are now down to
one equation with one variable, y.
Now solve for y.
Now solve for x by substituting the value of y
into either one of the original equations.
The solution for this system is x = 4 and
y = -2.
36
Draft
C. D. Pilmer
Example 2:
Solve the following system using substitution.
5 x + 2 y = −1 and 4 x − y = 7
Answer:
4x − y = 7
− y = −4 x + 7
y = 4x − 7
Choose one of the equations and isolate one
of the variables by expressing that variable in
terms of the other. In this example, we took
the second equation and isolated y. It was
the easiest thing to do.
5 x + 2 y = −1
5 x + 2(4 x − 7 ) = −1
5 x + 2(4 x − 7 ) = −1
5 x + 8 x − 14 = −1
5 x + 8 x = −1 + 14
13 x = 13
x =1
5 x + 2 y = −1
5(1) + 2 y = −1
The solution for this system is x = 1 and
y = -3.
y = −3
Answer:
NSSAL
©2009
Now solve for x.
Now solve for y by substituting the value of x
into either one of the original equations.
5 + 2 y = −1
2 y = −1 − 5
2 y = −6
Example 3:
Solve the system.
Substitute the expression for y into the other
equation. By doing this, we are now down to
one equation with one variable, x.
2 x + 3 y = 18 and 3 x − 4 y = −7
2 x + 3 y = 18
2 x = −3 y + 18
3
x = − y+9
2
3x − 4 y = −7
3 x − 4 y = −7

 3
3 − y + 9  − 4 y = −7

 2
9
− y + 27 − 4 y = −7
2
 9 
2 − y  + 2(27 ) − 2(4 y ) = 2(− 7 )
 2 
− 9 y + 54 − 8 y = −14
− 17 y = −14 − 54
− 17 y = −68
y=4
The answer is x = 3
and y = 4.
37
3 x − 4(4 ) = −7
3 x − 16 = −7
3 x = −7 + 16
3x = 9
x=3
Draft
C. D. Pilmer
Example 4:
Porter Pools charges $30 for a service call, plus $45 per hour for labour. Bailey Pools and Spas
charges $50 for a service call, plus $40 per hour for labour. Find the length of a service call for
which both companies charge the same amount.
Answer:
Determine the equations of the linear functions that describe the cost, c, of the service call in
terms of the time, t, to complete the jobs.
Porter Pools: c = 45t + 30
Bailey Pools and Spas: c = 40t + 50
Solve by substitution.
c = 45t + 30
40t + 50 = 45t + 30
40t − 45t = 30 − 50
− 5t = −20
− 5t − 20
=
−5
−5
t=4
We only need to solve for the time, t, for this question. Both companies charge the same
amount if it takes 4 hours to complete the service call.
Example 5:
The difference of two numbers is 5. Twice the larger number plus three times the smaller
number is 45. Find the numbers.
Answer:
Let x represent the larger number.
Let y represent the smaller number.
Determine the two equations.
x − y = 5 and 2 x + 3 y = 45
Solve by substitution.
x− y=5
2 x + 3 y = 45
x = y+5
2( y + 5) + 3 y = 45
2 y + 10 + 3 y = 45
2 y + 3 y = 45 − 10
5 y = 35
y=7
x− y=5
x−7 =5
x =5+7
x = 12
The larger number is 12. The smaller number is 7.
Note:
At the point of intersection, the x-coordinates for both graphs are equal, and the y-coordinates are
equal. The method of substitution uses this relationship to replace one variable with an
expression in terms of the other variable. This means that we go from two equations with two
unknown variables to one equation with one unknown variable.
NSSAL
©2009
38
Draft
C. D. Pilmer
Questions:
1. Solve the following systems by substitution.
(a) 4 x + y = 13 and 3 x + 5 y = 31
(b) 2 x + 6 y = 14 and x − 3 y = 1
(c) 2 x − y = 9 and 4 x + 5 y = 39
(d) 3 x + 4 y = 7 and − x + 3 y = −11
NSSAL
©2009
39
Draft
C. D. Pilmer
(e) y = 2 x − 5 and y = −3 x + 15
(f) 3 x + y = −4 and 5 x − 2 y = −25
(g) m − 4n = 4 and − 2m + 7n = −6
(h) 4r − s = −19 and 2r − 3s = 3
Scrambled Answer Key for Question 1
5, -2
-4, -2
2, 5
-1, 8
NSSAL
©2009
-3, 5
40
4, 1
-6, -5
6, 3
4, 3
Draft
C. D. Pilmer
2. Solve the following systems by substitution.
(a) x = 3 y + 5 and x = 5 y + 7
(b) 2 g + 4h = 10 and 6 g − 3h = −15
(c) 3 p − 2q = 14 and 2 p + 4q = 4
(d) 4 x − 3 y = −4 and 6 x + y = 5
NSSAL
©2009
41
Draft
C. D. Pilmer
3. Ajay is draining two different tanks using two different pumps. The first tank holds 85 litres
and is being drained by a pump that removes 7 litres per minute. The second tank holds 64
litres and is being drained by a pump that removes 4 litres per minute. Assuming that the
tanks were initially fully and that the pumps were turned on at the same time, at what time
will the tanks have the same number of litres of fluid?
4. Nita is going to organize a 50th wedding anniversary party for her parents. She wants to rent
a banquet hall and have a buffet. The local legion hall charges a flat rate of $185 for the hall,
plus $10 per person for the buffet. The voluntary fire department and its auxiliary charge a
flat rate of $215 for their hall and $8.50 per person for the buffet. Determine the number of
people who would have to attend in order for the two organizations to charge the same
amount.
5. The sum of two numbers is 12. If you double the first number and triple the second number,
then the new sum is 27. Find the numbers.
6. The difference of two numbers is 11. The larger number is five more than double the smaller
number. Find the numbers.
NSSAL
©2009
42
Draft
C. D. Pilmer
7. The perimeter of a rectangular garden is 174 metres. The length of the garden is twelve more
than twice the width. Find the length and width.
8. Two gears have a total of 75 teeth. The smaller gear has 15 more teeth
than half the number of teeth on the larger gear. How many teeth does
each gear have?
9. If Jacob buys 4 hamburgers and 1 container of french fries, he pays $15. If he purchases 6
hamburgers and 3 containers of french fries, he pays $24. How much does hamburger cost?
How much does a container of french fries cost?
10. Your parents have retired in the Florida Keys. They have to purchase an air conditioner for
their condominium. Unit A costs $650 to purchase and $18 per month to operate. Unit B
costs $574 to purchase and $22 per month to operate. Find the number of months at which
the total cost of either unit is the same.
NSSAL
©2009
43
Draft
C. D. Pilmer
Solving Systems Using Elimination
Up to this point we have learned how to solve systems of linear equations graphically and by
substitution. Although the method of substitution is vastly superior to the graphical method, it
can be problematic with some questions where neither of the equations have a coefficient that is
equal to 1 or -1. In many cases, it forces us to work with fractions.
Example: Solve the system. 3 x − 2 y = 7 and 4 x − 3 y = 8
Answer:
3x − 2 y = 7
4x − 3y = 8
− 2 y = −3 x + 7
− 2 y − 3x 7
=
+
−2
−2 −2
3
7
y= x−
2
2
3x − 2 y = 7
7
3
4 x − 3 x −  = 8
2
2
9
21
4x − x +
=8
2
2
 9   21 
2(4 x ) − 2 x  + 2  = 2(8)
2   2 
8 x − 9 x + 21 = 16
− 1x = −5
x=5
3(5) − 2 y = 7
15 − 2 y = 7
− 2 y = −8
y=4
The answer is x = 5
and y = 4.
Although we are capable of dealing with these types of questions, it would be nice if we could
find an easier method. Elimination is such a method. Elimination is an algebraic approach to
solving systems of linear equations in which two equations are added together in such a manner
that one of the unknown variables is eliminated. In many cases, to insure this elimination of a
variable, one or both of the equations must be altered before the two equations are added.
Example 1:
Solve the following system using elimination.
Answer:
3x − 2 y = 7
4x − 3 y = 8
×4
× -3
12 x − 8 y = 28
− 12 x + 9 y = −24
y=4
3x − 2 y = 7
3 x − 2(4) = 7
3 x = 15
x=5
NSSAL
©2009
3 x − 2 y = 7 and 4 x − 3 y = 8
Choose a variable to eliminate. We will eliminate
the x. Multiply each equation by a number that
gives the opposite coefficients for that variable in
both equations. If we multiply the first equation
by 4, the new coefficient of x will be 12. If we
multiply the second equation by -3, the new
coefficient of x will be -12. When we add the
two equations the 12x and -12x will cancel out
thus eliminating the variable x.
Now that we have solved for one variable, y, use
either of the two original equations to solve for
the other variable, x.
The answer is x = 5 and y = 4.
44
Draft
C. D. Pilmer
Example 2:
Solve the following system using elimination.
Answer:
5 x − 6 y = −16
2 x + 3 y = −1
×2
5 x − 6 y = −16
4x + 6 y = − 2
9 x = −18
x = −2
5 x − 6 y = −16 and 2 x + 3 y = −1
We will eliminate the y. We will leave the first
equation so that the coefficient of y will remain 6. If we multiply the second equation by 2, the
new coefficient of y will be 6. When we add the
two equations the -6y and 6y will cancel out thus
eliminating the variable y.
5 x − 6 y = −16
5(− 2) − 6 y = −16
− 10 − 6 y = −16
− 6 y = −6
Now that we have solved for one variable, x, use
either of the two original equations to solve for
the other variable, y.
y =1
The answer is x = -2 and y = 1.
Example 3:
Solve the following system using elimination.
Answer:
2x + 5 y = 4
x − 5 y = 17
3 x = 21
x=7
x − 5 y = 17
7 − 5 y = 17
− 5 y = 10
y = −2
2 x + 5 y = 4 and x − 5 y = 17
On a few rare occasions, neither of the equations
has to be altered to eliminate a variable. This is
the case with this particular question. The 5y and
-5y will cancel out when the two equations are
added together.
Now that we have solved for one variable, x, use
either of the two original equations to solve for
the other variable, y.
The answer is x = 7 and y = -2.
Example 4:
0.5 x + 0.7 y − 5 = 0 and
Solve the following system using elimination.
Answer:
0.5 x + 0.7 y − 5 = 0
0.5 x + 0.7 y = 5
2
1
x− y −3= 0
3
5
Both equations have to be rearranged so that they
are in the form Ax + By = C .
2
1
x − y −1 = 0
3
5
2
1
x − y =1
3
5
NSSAL
©2009
45
Draft
C. D. Pilmer
0.5 x + 0.7 y = 5
10(0.5 x ) + 10(0.7 y ) = 10(5)
5 x + 7 y = 50
It would be far easier to eliminate a variable if the
coefficients were integers, rather than decimals
and fractions.
We can get rid of the decimals in the first
equation by multiplying everything on both sides
of the equation by 10.
2
1
x − y =1
3
5
2 
1 
15 x  − 15 y  = 15(1)
3 
5 
10 x − 3 y = 15
We can get rid of the fractions in the second
equation by multiplying both sides of the
equation by 15; the least common multiple of
denominators 3 and 5.
5 x + 7 y = 50 × -2 − 10 x − 14 y = −100
10 x − 3 y = 15
10 x − 3 y = 15
Now that the equations are far easier to work
with, solve for x and y using elimination.
− 17 y = −85
y=5
5 x + 7 y = 50
5 x + 7(5) = 15
5 x + 35 = 50
5 x = 15
x=3
The answer is x = 3 and y = 5.
Example 5:
Tickets for a play cost $5 for adults and $3 for children. A total of 800 tickets were sold and the
total sales were $3600. How many adult tickets were sold?
Answer:
Let a represent the number of adult tickets sold
Let c represent the number of child ticket sold
Total Number Sold: 800 a + c = 800
Total Sales: $3600
5a + 3c = 3600
Now solve for a and c using elimination.
a + c = 800 × -3 − 3a − 3c = −2400
5a + 3c = 3600
NSSAL
©2009
5a + 3c = 3600
2a = 1200
a = 600
They sold 600 adult tickets.
46
Draft
C. D. Pilmer
Example 6:
Janice has a total of $2000 to invest. She puts part of it in an investment that pays 6% per year.
The remaining money she puts in an investment that pays 8% per year. At the end of the one
year, Janice has earned $134 in interest. How much money did she put in each of the
investments?
Answer:
Let A represent the amount she invested at 6% per year (Investment Plan A)
Let B represent the amount she invested at 8% per year (Investment Plan B)
Total Amount Invested: $2000
A + B = 2000
Interest Earned: $134
0.06 A + 0.08 B = 134
A + B = 2000
0.06 A + 0.08 B = 134
× −6
− 6 A − 6 B = −12000
× 100
6 A + 8 B = 13400
2 B = 1400
B = 700
A + B = 2000
A + 700 = 2000
A = 1300
Janice put $1300 in the investment that paid 6% interest per year. She put $700 in the
investment that paid 8% interest per year.
Example 7:
Andrew needs customized baseball hats with his company logo on them. He finds two local
companies that are able to produce the desired hats. Company A charges a $85 set-up fee, plus
$7.25 per hat. Company B charges a $120 set-up fee, plus $6 per hat. Under what circumstances
would Company A and Company B charge the same amount for the same number of hats?
Answer:
Find the equations of the two linear functions where c will represent the charges and n will
represent the number of hats that are produced.
Equation for Company A:
c = 7.25n + 85
Equation for Company B:
c = 6n + 120
Although this question is slightly easier to solve by substitution, we’ll use elimination. The
equations will have to be rearranged so that they are in the form Ax + By = C .
− 7.25n + c = 85
− 6n + c = 120
NSSAL
©2009
× -1
7.25n − c = −85
− 6n + c = 120
1.25n = 35
n = 28
The two companies would have
to manufacture 28 hats to keep
the costs the same.
47
Draft
C. D. Pilmer
Questions
1. Solve each of the systems using elimination.
(a) 4 x + 2 y = 18 and 3 x − 5 y = 20
(b) 6 x + 5 y = −19 and 5 x + 7 y = −13
(c) 3 g − 2h = 10 and 5 g − 9h = −6
(d) 2m + 7 n = 17 and 3m − 2n = −12
NSSAL
©2009
48
Draft
C. D. Pilmer
(e) 5 x − 2 y = −7 and 3 x + 4 y = 27
(f) 6 x + 7 y = −14 and 3 x − 5 y = 10
(g) 3 p + 8q = −39 and 7 p + 4q = −47
(h) 9 x + 5 y = 43 and 2 x − 5 y = 34
Scrambled Answer Key for Question 1
(0,-2)
(-4, 1)
(1, 6)
(3, -4)
NSSAL
©2009
(5, -1)
49
(-5, -3)
(6, 4)
(7, -4)
(-2, 3)
Draft
C. D. Pilmer
2. Solve the following systems using elimination.
(a) 2 y = 3 x + 2 and 4 y = 5 x + 8
(b) 3 p + 2q + 6 = 0 and 4 p − 5q + 31 = 0
(c) s = 3r + 15 and s = −2r + 5
(d) 3x + y = 20 and 0.2 x − 0.1 y = 1
NSSAL
©2009
50
Draft
C. D. Pilmer
(e) 0.7 g + 0.2h = 2 and g − 3h = 16
(f) 0.07 x + 0.06 y − 200 = 0 and 4 x − 5 y − 50 = 0
(g)
1
1
x + y = 1 and 2 x − 3 y = 22
4
2
(h)
1
1
p − q = −4 and p + 2q = 2
3
2
NSSAL
©2009
51
Draft
C. D. Pilmer
3. Marsha organized a fund raiser car wash for her son’s elementary school. They charged $7
to wash a car and $10 to wash a van. If they washed a total of 39 cars and vans, and earned
$321, how many cars were washed? How many vans were washed?
4. The sum of two numbers is 23. If the smaller number is doubled and then decreased by the
larger number, the answer is 4. Find the numbers.
5. An adult education instructor wants to purchase books from a second-hand bookstore for her
classroom. The softcover books cost $2. The hardcover books cost $5. If she buys 41 books
and pays $133, how many hardcover books did she purchase? How many softcover books
did she purchase?
6. Montez has to rent a truck to do some moving. Metro Truck Rental charges $60 per day plus
$0.18 per kilometre. U-Rent charges $75 for the day, plus $0.14 per kilometre. How far
would Montez have to travel with the truck in order that the charges from either rental
company would be the same?
NSSAL
©2009
52
Draft
C. D. Pilmer
7. Neil has $1100 to invest. Some of the money is placed in an account that pays 7% interest
per year. The remaining money is placed in an account that pays 5% interest per year. If
after one year, he has accumulated a total of $70.60 in interest, how much was invested in
each of these accounts?
8. Nasrin runs a catering business. She is catering a wedding where guests must choose
between a chicken dinner and a beef dinner. The chicken dinners cost $15 per plate, and the
beef dinners cost $18 per plate. The 54 wedding guests ordered in advance, and the total cost
to prepare the dinner is $879. How many of each type of dinner will Nasrin be preparing?
9. Heather is ordering shirts for her daughter’s canoe club. The short-sleeved shirts cost $5
each. The long-sleeved shirts cost $7 each. She ended up ordering 42 shirts and spending
$236. How many of each type of shirt were ordered?
NSSAL
©2009
53
Draft
C. D. Pilmer
Wrap-Up Statement:
In the last four sections you learned how to solve and check systems of equations questions
both in and out of context.
Reflect Upon Your Learning
Fill out this questionnaire after you have completed the four sections titled “Interpretation
Involving Two Linear Functions”, “Solving Systems of Equations Graphically”, “Solving
Systems by Substitution”, and “Solving Systems Using Elimination.” Select your response
to each statement.
1 - strongly disagree
2 - disagree
3 - neutral
4 - agree
5 - strongly agree
(a)
I understand all of the concepts covered in the section,
“Interpretation Involving Two Linear Functions.”
(b) I do not need any further assistance from the instructor on the
material covered in this section.
(c) I do not need any more practice questions.
(d) I understand all of the concepts covered in the section,
“Solving Systems of Equations Graphically.”
(e) I do not need any further assistance from the instructor on the
material covered in this section.
(f) I can solve for two unknowns given two equations by graphing
the linear functions and finding the point of intersection. I do
not need any more practice questions.
(g) I understand all of the concepts covered in the section,
“Solving Systems by Substitution.”
(h) I do not need any further assistance from the instructor on the
material covered in this section.
(i) I can solve for two unknowns given two equations using
substitution. I do not need any more practice questions.
(j) I can solve word problems that require one to use substitution.
I do not need any more practice questions.
(k) I understand all of the concepts covered in the section,
“Solving Systems Using Elimination.”
(l) I do not need any further assistance from the instructor on the
material covered in this section.
(m) I can solve for two unknowns given two equations using
elimination. I do not need any more practice questions.
(n) I can solve word problems that require one to use elimination.
I do not need any more practice questions.
NSSAL
©2009
54
1
2
3
4
5
1
2
3
4
5
1
1
2
2
3
3
4
4
5
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
Draft
C. D. Pilmer
Inconsistent and Equivalent Equations
Case 1: Inconsistent Equations
Ryan was asked to solve the following system of equations.
2 x + 4 y = 20 and 3 x + 6 y = 6
He completed the following work but was unable to solve for x or y.
2 x + 4 y = 20 × 3
6 x + 12 y = 60
3 x + 6 y = 6 × −2 − 6 x − 12 y = −12
0 = 48 ???
Questions Regarding Case 1
1. Did he make any mistakes in his calculations?
______
2. To check Ryan’s work, change both linear functions to their slope y-intercept forms
( y = mx + b) , and graph them on the same coordinate system.
7
6
5
4
3
2
1
0
-7
-6
-5
-4
-3
-2
-1
-1
0
1
2
3
4
5
6
7
-2
-3
-4
-5
-6
-7
3. How are the two lines oriented with respect to each other?
4. Do the lines ever intersect?
______
5. Is there no solution, one solution or an infinite number of solutions to this system of
equations? Explain.
NSSAL
©2009
55
Draft
C. D. Pilmer
Case 2: Equivalent Equations
Michelle was asked to solve the following system of equations.
0.2 x − 0.1 y = −0.3 and − 10 x + 5 y = 15
She completed the following work but was unable to solve for x or y.
0.2 x − 0.1 y = −0.3
× 50
− 10 x + 5 y = 15
10 x − 5 y = −15
− 10 x + 5 y = 15
0 = 0 ???
Questions Regarding Case 2
1. Did she make any mistakes in her calculations? ______
2. To check Michelle’s work, change both linear functions to their slope y-intercept forms
( y = mx + b) , and graph them on the same coordinate system.
7
6
5
4
3
2
1
0
-7
-6
-5
-4
-3
-2
-1
-1
0
1
2
3
4
5
6
7
-2
-3
-4
-5
-6
-7
3. How are the two lines oriented with respect to each other?
4. Do the lines ever intersect?
______
5. Is there no solution, one solution or an infinite number of solutions to this system of
equations? Explain.
NSSAL
©2009
56
Draft
C. D. Pilmer
Conclusions:
•
If two lines are parallel, they will never intersect. If you attempted to complete
substitution or elimination with these two equations, then you would end up with a false
statement like 0 = 48, as you did in Case 1. If a false statement occurs then you are
dealing with inconsistent equations and there is no solution for that system of equations.
•
If two lines are actually the same line, then they intersect at an infinite number of points.
If you attempted to complete substitution or elimination with these two equations, then
you would end up with the true statement 0 = 0, as you did in Case 2. If this true
statement occurs then you are dealing with equivalent equations and there are an infinite
number of solutions for that system of equations.
Questions:
1. Solve the following systems of equations.
(a) − 9 x − 6 y = −15 and 6 x + 4 y = 10
(b) 4 x + 3 y = 5 and 2 x − 5 y = 9
(c) 10 x − 4 y = 2 and − 25 x + 10 y = 17
(d) − 4 x + 10 y = −8 and 0.2 x − 0.5 y = 0.4
(e) 0.2 x + 0.5 y = 6 and 6 x + 15 y = 30
(f) 3 x + 4 y = −17 and 2 x − y = −4
2. For equation 4 x − 5 y = 7 , write an equation that would make the system inconsistent, and
another equation that would make the system equivalent.
NSSAL
©2009
57
Draft
C. D. Pilmer
Break Even Points
In the last few sections we learned how to solve systems of equations questions. These questions
had two unknowns and we solved for these unknowns using graphs, by substitution or by
elimination. In this section we will deal with a specific real world application that requires that
we are familiar with systems of equations.
Every business has expenses and revenue associated with a product that they are selling. All
businesses want their revenue to be greater than their expenses so that a profit can be made.
Profit = Revenue – Expenses
For example, if a company has revenues of $670 thousand and expenses of $210 thousand, then
we can say they make a profit of $460 thousand. When the expenses exceed the revenue, then
the company incurs a loss.
When no profit or loss is taking place, then the business is at its break even point. This occurs
when the revenue is equal to the expenses.
Break Even Point:
Revenue = Expenses
Businesses need to know their break even point (or points if they are dealing with more than one
product) so that they can understand when they will be making profits or losses. For example if
a company knows that it must produce and sell 30 000 light bulbs each month to break even,
then sales greater than 30 000 will produce profits while sales less 30 000 will produce losses.
With each of these break even point questions, we will create two equations (a revenue equation
and an expenses equation) and then find their point of intersection using substitution or
elimination. This point of intersection will be the break even point where revenue is equal to
expenses.
Example
Ryan has purchased a steel building where he will be manufacturing pewter Christmas
ornaments. His monthly fixed expenses of $2100 cover his mortgage on the building, heat, and
water. He also has variable expenses associated with the number of ornament he makes.
Obviously it will cost more in labor and materials if he produces more ornaments. Each
ornament costs him $4 to produce and ship to clientele. He sells the individual ornaments to his
clientele for $9. How many ornaments must he sell to break even?
`
Answer:
Create an expense equation. The fixed expenses are $2100. The variable expenses are $4
per ornament. Therefore we end up with the following expense equation.
E = 4n + 2100
Create the revenue equation. The revenue is equal to the selling price of an individual
ornament multiplied by the number of ornaments sold and shipped.
R = 9n
At the break even point the revenue equals the expenses. Take the two equations, set them
equal to each other, and solve for the unknown, n.
Revenue = Expenses
9n = 4n + 2100
NSSAL
©2009
58
Draft
C. D. Pilmer
9n − 4n = 4n + 2100 − 4n
5n = 2100
5n 2100
=
5
5
n = 420 ornaments
Ryan needs to produce and sell 420 pewter ornaments in order to break even.
7000
6000
Break Even Point
5000
amount of money
From a graphical perceptive, Ryan’s two
equations and break even point can be
represented in the following manner. We do
not need to produce graphs in order to solve
these types of questions.
4000
Expense Function
3000
2000
Revenue Function
1000
0
0
100
200
300
400
500
600
700
number of ornaments
Questions:
1. Nita hired a local manufacturer to produce customized T-shirts for a prestigious golf
tournament. The manufacturer is charging her $516 for designing the logo and programming
the stitching machine. He is also charging her $9.25 for each T-shirt. She will be selling
each shirt for $20.
(a) How many will she have to sell to break even?
(b) How much profit will she make if she sells 90 shirts?
(c) How much will she lose if she only sells 15 shirts?
NSSAL
©2009
59
Draft
C. D. Pilmer
2. A promoter is trying to arrange a concert on the Halifax Commons. He approached an
established rock band. The band informed him that they charge a flat rate of $667 000 and
take $20 for each ticket sold. The promoter must cover these costs and the city is prepared to
cover the other costs. The promoter can sell each ticket for $49.
(a) How many tickets must be sold to break even?
(b) How much will the promoter lose if he only sells 15 000 tickets?
(c) How much will the promoter make if he sells 40 000 tickets?
3. Kimi makes and bottles homemade jam that she sells to several specialty stores in Atlantic
Canada. She sells the jam to the stores for $3.20 each. Her fixed costs are $324 per month
and the variable cost per jar is $1.40.
(a) How many jars must she sell per month to break even?
(b) What will her profit be if she sells 300 jars in a month?
(c) How much will she lose if she only sells 110 jars in a month?
NSSAL
©2009
60
Draft
C. D. Pilmer
Putting It Together
Complete the following review questions.
Questions
1. Change each of the following linear functions to their explicit forms (i.e. “y =” form).
(a) 5 x + 6 y − 18 = 0
(b) 3 x − 7 y + 28 = y − 4
(c) 3(4 y + x ) = x − 24
(d) 5(2 x − y ) = 3 x + 45
(e) 0.3 x + 0.4 y − 5 = 1
(f) 0.2 x − 0.5 y = x + 1
(g)
5
2
y + x − 10 = 0
6
3
(h)
3
1
x − y + 19 = 4
5
4
2. Determine the x-intercept and y-intercept of the function 5 x − 6 y − 60 = 0 .
NSSAL
©2009
61
Draft
C. D. Pilmer
3. (a) Determine the equation of the linear function that passes through the points (20, 19) and
(5, 7).
(b) Determine the equation of the straight line that passes through (8, -17) and (-4, 10).
(c) Determine the equation of the linear function that has a x-intercept of -4 and passes
through (6, -15).
(d) Determine the equation of the linear function that has a x-intercept of 9 and a y-intercept
of -15.
(e) Determine the x-intercept of the linear function that passes through (-15, 6) and (5, 18).
NSSAL
©2009
62
Draft
C. D. Pilmer
4. Elliot wants to buy gravel from one particular supplier. There is a trucking fee plus a cost
per tonne of gravel. He knows that if he orders 8 tonnes, he will be charged $234. If he
orders 13 tonnes, he will be charged $349.
(a) Determine the equation of the linear function that describes the cost, C, of the gravel in
terms of the number, n, of tonnes.
(b) How much does it cost to order 15 tonnes of gravel?
(c) How much is the trucking fee?
(d) How many tones of travel can he order with $280.
(e) If the trucking fee was $45 and the cost per tonne of gravel was $22, what would be the
equation of the linear function describing the cost in terms of the number of tones?
3
5. Given the linear function y = − x + 10 , which of the listed functions below are:
7
(a) parallel to it?
_________________
(b) perpendicular to it?
_________________
Listed Functions:
2
(i) y = x + 10
5
3
(ii) y = − x − 8
7
(iii) y =
3
1
(iv) y = − x −
7
10
9
1
(v) y = − x −
4
10
7
(vi) y = − x + 10
3
7
(viii) y = − x − 6
3
(ix) y =
(vii) y =
NSSAL
©2009
3
x + 10
7
63
7
x+2
3
7
x
3
Draft
C. D. Pilmer
6. Solve the following system using all three techniques.
7 x + 4 y = 24 and x − 2 y = 6
(a) Graphically
7
6
5
4
3
2
1
0
-7
-6
-5
-4
-3
-2
-1
-1
0
1
2
3
4
5
6
7
-2
-3
-4
-5
-6
-7
(b) Substitution
(c) Elimination
For the questions 7 to 14, solve using an algebraic technique (i.e. substitution or elimination).
7. The sum of two numbers is 19. If you double the smaller number and then decrease it by the
larger number, the result is 5. Find the numbers.
8. The two numbers differ by 8. If you double the larger number and then increase this by triple
the smaller number, the result is 81. Find the numbers.
NSSAL
©2009
64
Draft
C. D. Pilmer
9. Kiana is buying canned peas and canned corn for the local food bank. The peas cost $0.89
per can. The corn costs $1.19 per can. She bought 75 cans and paid $80.25. How many of
each type did she purchase?
10. If Thomas buys 6 hamburgers and 2 chili dogs, he pays $21. If he buys 4 hamburgers and 8
chili dogs, he pays $29. How much does a hamburger cost? How much does a chili dog
cost?
11. Two very large containers, that initially have some water in them, are being filled using
different pumps. The first container initially has 14 litres of water in it and is being filled at a
rate of 1.5 litres per minute. The second container initially has 5 litres of water in it and is
being filled at a rate of 2.1 litres per minute. At what time will the two containers have the
same amount of water?
12. The perimeter of a rectangle is 68 cm. The length of the rectangle is one centimeter more
than double the width. Find the length and width of the rectangle.
NSSAL
©2009
65
Draft
C. D. Pilmer
13. Monique has $850 to invest. Some of the money is placed in an account that pays 7%
interest per year. The remaining money is placed in an account that pays 6% interest per
year. If after one year, she has accumulated a total of $57.10 in interest, how much was
invested in each of these accounts?
14. Solve the following systems algebraically.
(a) 2 x + 3 y = 4 and 0.8 x + 1.2 y = 3
(b) 9 x = 6 y + 15 and
1
1
5
x− y=
2
3
6
15. Monique is trying to raise money for her son’s scout troop. She decides to host a supper at a
local fire hall. It will cost her $310 to rent the hall for the evening and $4.25 to make each
meal. She decides to charge $12 for each meal.
(a) How many meals does she have to sell to break even?
(b) How much profit will she make if she sells 70 meals?
NSSAL
©2009
66
Draft
C. D. Pilmer
Wrap-Up Statement:
In the last three sections you learned about equivalent and inconsistent equations, break even
points, and reviewed the concepts addressed in the previous four sections.
Reflect Upon Your Learning
Fill out this questionnaire after you have completed the two sections titled “Inconsistent and
Equivalent Equations”, “Break Even Points” and “Putting It Together.” Select your
response to each statement.
1 - strongly disagree
2 - disagree
3 - neutral
4 - agree
5 - strongly agree
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
NSSAL
©2009
I understand all of the concepts covered in the section,
“Inconsistent and Equivalent Equations.”
I understand that inconsistent equations are formed when two
linear functions are parallel. There is no solution to this system
of equations because the lines never intersect.
I understand that equivalent equations are formed when two
equations actually represent the same linear function. There
are an infinite number of solutions because the two equations
intersect each other along their entire length.
I do not need any further assistance from the instructor on the
material covered in this section.
I do not need any more practice questions on the material
covered in this section.
I understand all of the concepts covered in the section, “Break
Even Points.”
I do not need any further assistance from the instructor on the
material covered in this section.
I understand all of the concepts covered in the review section,
“Putting It Together.”
After completing the review, I feel confident about the material
I learned over the last few weeks.
67
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
Draft
C. D. Pilmer
Post-Unit Reflections
What is the most valuable or important
thing you learned in this unit?
What part did you find most interesting or
enjoyable?
What was the most challenging part, and
how did you respond to this challenge?
How did you feel about this math topic
when you started this unit?
How do you feel about this math topic
now?
Of the skills you used in this unit, which
is your strongest skill?
What skill(s) do you feel you need to
improve, and how will you improve them?
How does what you learned in this unit fit
with your personal goals?
NSSAL
©2009
68
Draft
C. D. Pilmer
Additional Practice: Explicit Form
Change each of the following to their explicit form (i.e. “y =” form) and then determine if the
function is linear.
1. 3 x − 2 y + 10 = 0
2. 9 x + 3 y − 2 = 5 y + x
3. 3( y + 2 x ) = 6 x + 12
4. 4(10 − 2 x ) = 5( y + 4 )
5. 0.4 y + 1.2 x + 1 = −1
6. 0.3 x 2 − 2 y = 2.4
7.
1
3
x + y + 4 = −2
3
4
NSSAL
©2009
(
8.
69
)
1
4
y+
x −3=5
2
5
Draft
C. D. Pilmer
Answers
Introduction to Linear Functions (pages 1 to 8)
1. Graphs (a), (d), and (h)
1
1
2. (a) y = − x + 2 , slope = − , y-intercept = 2
3
3
9
9
(b) y = x + 7 , slope = , y-intercept = 7
2
2
(c) y = 2 x − 3 , slope = 2, y-intercept = -3
3
3
(d) y = x − 7 , slope = , y-intercept = -7
2
2
3
3
(e) y =
, y-intercept = -2
x − 2 , slope =
10
10
1
1
(f) y = x − 2 , slope = , y-intercept = -2
3
3
5
5
(g) y = − x + 6 , slope = − , y-intercept = 6
2
2
19
19
(h) y = x − 4 , slope =
, y-intercept = -4
6
6
7
7
(i) y = − x + 80 , slope = − , y-intercept = 80
6
6
20
20
(j) y = −
, y-intercept = 0
x , slope = −
3
3
5
5
(k) y = − x + 75 , slope = − , y-intercept = 75
6
6
15
15
(l) y = − x + 144 , slope = − , y-intercept = 144
2
2
3. Equations (a), (b), (e), (f), (h), (j), and (l)
4. (a) y = −2 x +
(c) y = −
17
, linear
2
1
x + 2 , linear
2
(b) y =
5
x − 2 , linear
3
(d) y = 2 x + 7 , not linear
3
x + 9 , linear
4
(e) y = 4 , not linear
(f) y =
3
x + 2 , linear
7
7
(i) y =
+ 4 , not linear
2x
(h) x = −1 , not linear
(g) y = −
NSSAL
©2009
(j) y = −7 x 2 − 3 x + 2 , not linear
70
Draft
C. D. Pilmer
(k) y =
5.
25
29
, linear
x+
3
3
(l) y = 6 x 3 − 12 x + 6 , not linear
Equation
1
y =− x+4
3
Line
y = −2 x − 3
E
C
1
x−3
2
4
y = x+4
3
y=
D
A
y = 3x
B
6.
7
6
5
4
3
(b)
2
1
0
-7
-6
-5
-4
-3
-2
-1
-1
0
1
2
3
4
5
6
7
-2
(a)
-3
-4
-5
-6
-7
7. (a)
(b)
50
7
6
40
5
4
30
3
20
2
10
1
0
0
-100
-75
-50
-25
-1 0
25
50
75
100
-4
-3
-2
-1
0
1
2
3
4
-10
-2
-3
-20
-4
-30
-5
-40
-6
-7
-50
Find the Equation Given Points and/or Intercepts (pages 9 to 19)
1. y = −2 x + 8
NSSAL
©2009
2.
y=
1
x−5
2
3.
71
5
y = − x −1
3
Draft
C. D. Pilmer
3
x − 24
4
5.
2
y =− x+4
7
4
7. y = − x − 12
3
8.
y=
5
10. y = − x + 6
8
9
11. y = − x + 2
7
4. y =
3
x+7
2
6.
y = −5 x + 8
9.
y=
7
x − 14
5
12. y =
8
x − 16
9
3
13. y = − x − 9
7
14.
(a)
(b)
(c)
(d)
x-intercept
15
6
-4
-12
9
x+6
4
15. y =
y-intercept
6
-8
14
10
7
16. y = − x + 14
2
17. More information was supplied than needed.
18. x-intercept: 24
19. x-intercept: 6
Applications of Linear Functions (pages 20 to 27)
1. (a)
(b)
(c)
(d)
(e)
(f)
(g)
a = −80t + 1000
600 mL
1000 ml
10.5 hours
12.5 hours
slope = -80; The I.V. bag is draining at a rate of 80 mL per hour
a = −60t + 500
40
t
3
267 m
No, he's probably not going to be traveling at a constant speed for the next 10 minutes.
27 seconds
The slope is positive because the distance between Markus and the cyclist is increasing
with time.
2. (a) d =
(b)
(c)
(d)
(e)
NSSAL
©2009
72
Draft
C. D. Pilmer
(f) d =
83
t
6
e = 28h + 560
$560
11 hours
$28 per hour
Positive. This makes sense because her earnings should increase as she works more
hours.
(f) The flat rate for 40 hours of work is $360 and she receives $18 per hour for overtime
work.
3. (a)
(b)
(c)
(d)
(e)
4. (a)
(b)
(c)
(d)
(e)
(f)
(g)
M = −0.7t + 5.8
0.7 kg of pellets per hour
3.7 kg
5.8 kg
7 hours
8.3 hours
M = −0.6t + 6.3
5. (a) F = 1.8C + 32
(b) 93.2oF
(c) 18.3oC
Parallel and Perpendicular Lines (pages 28 to 29)
1. The lines are parallel.
2. The slopes are equal. We can conclude that parallel lines have equal slopes.
3. Infinite Number of Answers: … y =
4
4
4
4
x − 1 , y = x , y = x + 1 , y = x + 2 ,…
7
7
7
7
4. The lines are perpendicular to each other.
5. If we take the reciprocal (i.e. interchange the top and bottom of the fraction) of one of the
slopes and then change the sign, we end up with the slope of the other line. We can conclude
that perpendicular lines have slopes that are negative reciprocals of one another.
6. Infinite Number of Answers: … y = −
7. Infinite Number of Answers: … y =
NSSAL
©2009
5
5
5
5
x − 1 , y = − x , y = − x + 1 , y = − x + 2 ,…
2
2
2
2
3
3
3
3
x − 1 , y = x , y = x + 1 , y = x + 2 ,…
8
8
8
8
73
Draft
C. D. Pilmer
8. (a) parallel: iii and iv
(b) perpendicular: i and v
Solving Systems of Equations Graphically (pages 30 to 35)
1. (a) slope = −
2
, y-intercept = 5
3
4
, y-intercept = 2
3
(c) slope = 4, y-intercept = -5
(b) slope =
2. (a) solution
(b) not solution
(c) solution
3. (a) x = 3, y = 2
(c) x = -5, y = -1
(e) x = 5, y = -4
4. (a)
(b)
(c)
(d)
(b) x = 4, y = -3
(d) x = -2, y = 3
c = 30h + 150
c = 25h + 165
(3, 240)
If both painters work for 3 hours, they charge the same amount ($240).
Solving Systems by Substitution (pages 36 to 43)
1. (a)
(c)
(e)
(g)
x = 2, y = 5
x = 6, y = 3
x = 4, y = 3
m = -4, n = -2
(b)
(d)
(f)
(h)
2. (a) x = 2, y = -1
(c) p = 4, q = -1
x = 4, y = 1
x = 5, y = -2
x = -3, y = 5
r = -6, s = -5
(b) g = -1, h = 3
(d) x = 0.5, y = 2
3. v = −7t + 85 , v = −4t + 64 , 7 minutes
4. c = 10n + 185 , c = 8.5n + 215 , 20 people
5. x + y = 12 , 2 x + 3 y = 27 , The first number is 9. The second number is 3.
6. x − y = 11 , x = 2 y + 5 , The larger number is 17. The smaller number is 6.
7. 2l + 2 w = 174 , l = 2w + 12 , The length is 62 metres. The width is 25 metres.
8. s + l = 75 , s = 0.5l + 15 , The larger gear has 40 teeth. The smaller gear has 35 teeth.
NSSAL
©2009
74
Draft
C. D. Pilmer
9. 4h + 1 f = 15 , 6h + 3 f = 24 , Hamburger: $3.50
French Fries: $1
10. c = 18m + 650 , c = 22n + 574 , 19 months
Solving Systems Using Elimination (pages 44 to 54)
1. (a) x = 5, y = -1
(b) x = -4, y = 1
(c) g = 6, h = 4
(d) m = -2, n = 3
(e) x = 1, y = 6
(f) x = 0, y = -2
(g) p = -5, q = -3
(h) x = 7, y = -4
2. (a) x = 4, y = 7
(b) p = -4, q = 3
(c) r = -2, s = 9
(d) x = 6, y = 2
(e) g = 4, h = -4
(f) x = 1700, y = 1350
(g) x = 8, y = -2
(h) p = -6, q = 4
3. c + v = 39 , 7c + 10v = 321 , 23 cars and 16 vans
4. s + l = 23 , 2 s − l = 4 , 14 and 9
5. s + h = 41 , 2 s + 5h = 133 , 17 hardcover books and 24 softcover books
6. c = 0.18d + 60 , c = 0.14d + 75 , 375 kilometres
7. x + y = 1100 , 0.07 x + 0.05 y = 70.60
$780 was invested in the account that pays 7% interest per year
$320 was invested in the account that pays 5% interest per year
8. c + b = 54 , 15c + 18b = 879
31 chicken dinners and 23 beef dinners
9. s + l = 42 , 5s + 7l = 236
29 short-sleeved shirts and 13 long-sleeved shirts
NSSAL
©2009
75
Draft
C. D. Pilmer
Inconsistent and Equivalent Equations (pages 55 to 57)
Case 1
1. No
1
1
2. y = − x + 5 and y = − x + 1
2
2
3. The lines are parallel to each other.
4. No
5. There is no solution because the two lines never intersect.
Case 2
1. No
2. y = 2 x + 3 and y = 2 x + 3
3. The two lines lie right on top of each other. They are actually the same line.
4. Yes
5. There are an infinite number of solutions because the two lines lie right on top of each other
(infinite number of points of intersection).
Questions
1. (a)
(b)
(c)
(d)
(e)
(f)
infinite number of solutions (equivalent equations)
x = 2 and y = -1
no solution (inconsistent equations)
infinite number of solutions (equivalent equations)
no solution (inconsistent equations)
x = -3 and y = -2
2. Answers will vary.
Break Even Points (pages 58 to 60)
1. (a) 48 T-shirts
(b) Hint: Work out the expenses and revenue corresponding to 90 shirts and then take the
difference between the two.
Answer: $451.50
(c) $354.75
2. (a) 23 000 tickets
(b) $232 000
(c) $493 000
NSSAL
©2009
76
Draft
C. D. Pilmer
3. (a) 180 jars
(b) $216
(c) $126
Putting It Together (pages 61 to 67)
5
1. (a) y = − x + 3
6
1
(c) y = − x − 2
6
3
(e) y = − x + 15
4
4
(g) y = − x + 12
5
2. x-intercept = 12
3
(b) y = x + 4
8
7
(d) y = x − 9
5
8
(f) y = − x − 2
5
12
(h) y = x + 60
5
y-intercept = -10
3. (a) y =
9
(b) y = − x + 1
4
5
(d) y = x − 15
3
4. (a) C = 23n + 50
(c) $50
(e) C = 22n + 45
(b) $395
(d) 10 tonnes
5. (a) parallel: (ii) and (iv)
(b) perpendicular: (iii) and (ix)
4
x+3
5
3
(c) y = − x − 6
2
3
(e) y = x + 15 , x-intercept = -25
5
6. x = 4 and y = -1
7. x + y = 19 , 2 x − y = 5 , The larger number is 11. The smaller number is 8.
8. x − y = 8 , 2 x + 3 y = 81 , The larger number is 21. The smaller number is 13.
9.
p + c = 75 , 0.89 p + 1.19c = 80.25 , 30 cans of peas and 45 cans of corn
10. 6h + 2d = 21 , 4h + 8d = 29 , Hamburger: $2.75, Chili Dog: $2.25
11. v = 1.5t + 14 , v = 2.1t + 5 , 15 minutes
12. 2l + 2 w = 68 , l = 2 w + 1 , Length: 23 cm, Width: 11 cm
NSSAL
©2009
77
Draft
C. D. Pilmer
13. x + y = 850 , 0.07 x + 0.06 y = 57.1
$610 was invested in the account that pays 7% interest per year
$240 was invested in the account that pays 6% interest per year
14. (a) no solution (inconsistent equations, parallel lines)
(b) infinite number of solutions (equivalent equations)
15. (a) 40 meals
(b) $232.50
Additional Practice: Explicit Form (page 69)
1.
y=
3.
3
x + 5 , linear
2
2.
y = 4 x − 1 , linear
y = 4 , not linear
4.
8
y = − x + 4 , linear
5
5.
y = −3 x − 5 , linear
6.
y=
7.
y=−
8.
y=−
NSSAL
©2009
4
x − 8 , linear
9
78
1 2
x − 4 , not linear
2
8
x + 16 , not linear
5
Draft
C. D. Pilmer