Algebra
Module A50
Systems of
Equations
Copyright
This publication © The Northern
Alberta Institute of Technology
2002. All Rights Reserved.
LAST REVISED Nov, 2008
Systems of Equations
Statement of Prerequisite Skills
Complete all previous TLM modules before beginning this module.
Required Supporting Materials
Access to the World Wide Web.
Internet Explorer 5.5 or greater.
Macromedia Flash Player.
Rationale
Why is it important for you to learn this material?
Systems of equations are used to solve many problems in engineering, science, business
and other fields. This module will give you practice in using systems of equations to
solve problems and in translating problems from English into the language of math.
Learning Outcome
When you complete this module you will be able to…
Solve systems of linear equations.
Learning Objectives
1.
2.
3.
4.
5.
6.
Solve simultaneous equations graphically.
Solve simultaneous equations using the method of addition and/or subtraction.
Solve simultaneous equations using the method of substitution.
Solve a system of three unknowns using the method of elimination.
Solve word problems requiring the solution of 2 by 2 systems of equations.
Solve word problems requiring the solution of 3 by 3 systems of equations.
1
Module A50 − Systems of Equations
Connection Activity
To complete this Learning Outcome Guide, you will consider how you would solve the
following problem.
A laboratory technician wished to make 150L of 55% alcohol solution by mixing 45%
alcohol and 70% alcohol solutions. Determine the number of litres of each solution that
should be used.
The skills learned in this outcome guide will allow you to solve problems similar to this
in the world of work.
Answer: See objective 5 exercises.
2
Module A50 − Systems of Equations
OBJECTIVE ONE
When you complete this objective you will be able to…
Solve simultaneous equations graphically.
Exploration Activity
When you solve a system of two linear equations graphically you are determining
whether two lines:
1. Intersect OR
2. Do not intersect.
THERE ARE THREE POSSIBLE CHOICES.
y
y
y
l2
A
x
x
l1
x
l1
l2
one point of intersection at A
many points of intersection since l1 and
l2 are the same line
no points of intersection since l1 and l2
are parallel
Figure 1A
Figure 1B
Figure 1C
General Names of Solutions to Systems of Equations
1. Consistent solution
- the point of intersection is the solution to the system of equations
- one point of intersection
- Fig. 1A
2. Dependent solution
- no unique solution to the system of equations.
- many points of intersection
- Fig. 1B
3. Inconsistent solution
- no solution to the system of equations
- no points of intersection
- Fig. 1C
3
Module A50 − Systems of Equations
Graphical Solution to a System
To find the graphical solution to a system of equations we draw the graph for each of the
given algebraic equations on the same coordinate axis. Then we must identify the
ordered pair that is determined at the point of intersection because this is the solution to
the system of equations.
EXAMPLE 1
Solve the following system graphically.
2x + 3y = −12 (1)
2x – y = −4
(2)
SOLUTION:
Sketch the graphs of the given equations using the x and y intercept method.
2x + 3y = −12
2x – y = −4
x – intercept
y – intercept
x – intercept
y- intercept
set y = 0
set x = 0
set y = 0
set x = 0
2x + 3(0) = −12
2(0) + 3y = −12
2x – 0 = −4
2(0) – y = −4
2x = −12
3y = −12
2x = −4
−y = −4
x = −6
y = −4
x = −2
y=4
(−6, 0)
(0, −4)
(−2, 0)
(0, 4)
6
2x - y = -4
4
2
-6
-4
2
-2
4
6
x
-2
Solution (-3,-2)
-4
2x + 3y = -2
-6
4
Module A50 − Systems of Equations
NOTES:
1. If the solution to a linear system is a dependent solution then state: dependent
solution.
2. If the solution to a linear system is an inconsistent solution then state:
inconsistent solution.
3.
Check your solution to a system of equations using the same procedure that you
used in earlier modules on equations.
CHECK OF PREVIOUS EXAMPLE:
Equations
2x + 3y = −12 (1)
2x – y = −4
(2)
SOLUTION:
from the graph above we have the point of intersection at:
x = −3
y = −2
2x + 3y =
−12
2x – y =
2(−3) + 3(−2)
2(−3) – (−2)
−6 − 6
−6 + 2
−12
−4
Therefore LHS = RHS
Therefore LHS = RHS
−4
It is often difficult to get an exact solution to a system of equations by graphing as the
point of intersection is often decimal numbers. In the following objectives we will see
how to find the exact solution algebraically.
5
Module A50 − Systems of Equations
Experiential Activity One
Solve the following systems graphically. Check your solutions.
1. 3x – 2y = 5
x + 2y = −3
2. 2x – y = −1
−5x + 3y = −1
3. x = 4
2x + 3y = 5
4. 3x – 5y = 4
−6x + 10y = −8
5. 2x – 3y = −15
2x – 3y = 6
Show Me.
Visit the following web page for an excellent activity that allows you to experiment with
two by two systems and their graphical solutions:
http://www.exploremath.com/activities/Activity_page.cfm?ActivityID=18
Experiential Activity One Answers
Check you answers by substituting the variable solutions into the original equations. The
solutions should work for all equations in the system.
6
Module A50 − Systems of Equations
OBJECTIVE TWO
When you complete this objective you will be able to…
Solve simultaneous equations using the method of addition and/or subtraction.
Exploration Activity
EXAMPLE 1
Solve the following system algebraically by the method of addition or subtraction.
2x + 3y = −12 (equation 1)
2x – y = −4
(equation 2)
Same example as in Objective 1.
SOLUTION:
NOTE:
This is a nice question because the x coefficient in both equations (1) and (2) is the same.
The idea is to eliminate either the x’s or the y’s (this time the x’s).
Subtract equation 2 from equation 1,
2x + 3y = −12 (equation 1)
−(2x – y = −4) (equation 2), Åsubtracting equation 2 from equation1
4y = −8
now solve for y
y = −2
** When we subtracted equation 2 from equation 1 the x variable was eliminated.
This was because the coefficients were the same. If the coefficients were
different then we would have had to do something to the equation(s) to make
them the same before subtracting.
Substitute y = −2 into equation 1 above to get x:
2x + 3(−2) = −12
2x – 6 = −12
2x = −6
x = −3
The solution is (−3, −2)
Check the solution. This is left to the student as an exercise.
Same as Objective 1, example 1.
7
Module A50 − Systems of Equations
EXAMPLE 2
Solve the following system algebraically by the method of addition or subtraction.
x + y = −3
3x – 2y = −4
equation 1
equation 2
SOLUTION:
Make the either the x or y coefficients the same so that when we add or subtract equations
one of the variables will be eliminated. In this case we will choose to eliminate the y’s
from equations 1 and 2 so we want the y coefficients to both be 2. We could have also
chosen to eliminate the x’s in which case we would want the x coefficients to be 3.
2(x + y = −3)
2x + 2y = −6
Multiply each term in equation 1 by 2 to make the y coefficient the same
as the y coefficient the same as in equation 2. This gives equation 3
Now we add the equations. We add the equations because they variables have opposite
signs.
2x + 2y = −6
+(3x – 2y = −4)
5x + 0 = −10
x = −2
equation 3
equation 2
Substitute x = −2 into equation 1 above to get y
To solve for y
−2 + y = −3
y = −1
Solution is (−2, −1)
Check the solution.
x+y=
−3
3x – 2y =
−2 − 1
3(−2) – 2(−1)
−3
−6 + 2
−4
−4
Therefore LHS = RHS
Therefore LHS = RHS
8
Module A50 − Systems of Equations
EXAMPLE 3
Solve by addition or subtraction.
5x + 3y = 1
10x + 6y = 0
equation 1
equation 2
SOLUTION:
Make the x coefficients the same in equations 1 and 2.
2(5x + 3y = 1)
10x + 6y = 2
multiply equation 1 by 2 to obtain equation 3
equation 3
Now subtract equation 2 from equation 3
10x + 6y = 2
−(10x + 6y = 0)
0+0=2
0=2
equation 3
equation 2
equation 3 − equation 2
When this happens the system has no point of intersection.
Answer: inconsistent solution
EXAMPLE 4
Solve the following system by addition or subtraction.
3x + y = 2
2 1
x= − y
3 3
equation 1
equation 2
SOLUTION:
Rewrite equation 2 in standard form by clearing the fractions to obtain equation 3
3x + y = 2
equation 3
Now subtract equation 3 from equation 1
3x + y = 2
−(3x + y = 2)
0+0=0
0=0
equation 1
equation 3
When this happens there are many points in common.
Answer: dependent solution
9
Module A50 − Systems of Equations
EXAMPLE 5
Solve the following system by addition or subtraction.
5x + 2y = 5
2x + 3y = 13
equation 1
equation 2
SOLUTION:
Make the coefficient in front of the x’s the same:
10x + 4y = 10
10x + 15y = 65
equation 3
equation 4
equation 1 × 2
equation 1 × 5
Subtract equation 4 from equation 3.
10x + 4y = 10
−(10x + 15y = 65)
−11y = −55
y=5
Now substitute y = 5 into equation 1 and solve for x,
5x + 2(5) = 5
5x + 10 = 5
5x = −5
x = −1
Solution is (−1,5)
Check the solution.
5x + 2y =
5
2x + 3y =
5(−1) + 2(5)
2(−1) + 3(5) =
−5 + 10
−2 + 15
5
Therefore LHS = RHS
=5
13
13
= 13
Therefore LHS = RHS
10
Module A50 − Systems of Equations
Experiential Activity Two
Solve the following systems algebraically by the method of addition or subtraction.
Check your solutions.
1. x + 5y = 4
x + 3y = 2
2. 3x + 2y = 34
5x – 3y = −13
3. 3(x + y) = 1 Show Me
3y = 2 – 3x
4. 3(x + 2y) = 4(y – x)
2y = 7x
5. y = 2 – 3x
x=
2 1
− y
3 3
Experiential Activity Two Answers
Check your answers by substituting the variable solutions into the original equations.
The solutions should work for all equations in the system.
1. x = −1
y=1
2. x = 4
y = 11
3. inconsistent solution
4. x = 0
y=0
5. dependent solution
11
Module A50 − Systems of Equations
OBJECTIVE THREE
When you complete this objective you will be able to…
Solve simultaneous equations using the method of substitution.
Exploration Activity
Note: This method is used in later modules when solving more complicated systems of
equations. These questions could be done using addition or subtraction but questions in
later modules will require the substitution method.
The method of substitution requires isolating a variable in one equation and then
substituting into the other equation and solving.
EXAMPLE 1
Solve the following system algebraically by the method of substitution. Check your
solution.
2x + 3y = −12 equation 1
2x − y = −4
equation 2
SOLUTION:
Isolate an x or a y from one of the above equations.
From equation 2 above solve for y:
2x − y = −4
−y = −4 – 2x
y = 4 + 2x
Substitute y = 4 + 2x into equation 1
To find the other unknown always substitute into the original equation you did not use.
Now solve for x:
2x + 3(4 + 2x) = −12
2x + 12 + 6x = −12
8x = −24
x = −3
12
Module A50 − Systems of Equations
Substitute x = −3 into equation 1 to find y:
2(−3) + 3y = −12
−6 + 3y = −12
3y = −6
y = −2
Solution is (−3, −2)
Check your solution
It is the same as the objective 1 example.
EXAMPLE 2
Solve the following system algebraically by the method of substitution. Check your
solution.
x y
=
equation 1
2 3
y–x=1
equation 2
SOLUTION:
Solving equation 2 for y we get:
y=x+1
Substitute y = x + 1 into equation 1
x x +1
=
, now cross-multiply to get:
2
3
3x = 2(x + 1)
3x = 2x + 2
x=2
Substitute x = 2 into equation 1
2 y
=
2 3
y=3
Solution is (2, 3)
Check your solution
13
Module A50 − Systems of Equations
Experiential Activity Three
1. 2x + y = 3
2y – 3x = 6
2.
1
1
=
x + 4y 2
Show Me.
8y – x = 7
3. 5x – 2y = −1
x + y = −3
4. 4x + 3y = 0
y–x= −
7
12
5. 10x + 10y = −3
y – x = 0.5
Experiential Activity Three Answers
Check your answers by substituting the variable solutions into the original equations.
The solutions should work for all equations in the system.
1. x = 0
y=3
2. x = −1
y=3
4
3. x = −1
y = −2
4.
x= 1
4
y=−1
3
14
Module A50 − Systems of Equations
OBJECTIVE FOUR
When you complete this objective you will be able to…
Solve a system of three unknowns using the method of addition/subtraction.
Exploration Activity
Solving a system of 3 equations using the method addition/subtraction requires
eliminating two variables. First, eliminate one variable to reduce to a system of 2
variables. Then use the methods already developed for 2 unknowns to continue solving.
Remember you must solve for all three variables!!
EXAMPLE 1
Solve the following system algebraically using the method of addition or subtraction.
3x – 2y + 4z = 11
2x + 3y – z = 5
x + 4y – 2z = 3
equation 1
equation 2
equation 3
SOLUTION:
Start by deciding which variable to eliminate first. Generally choose the variable which
is “easiest” to eliminate, but there is no rule to govern your choice. We chose the letter y.
Eliminate y from equation 1and equation 2
9x – 6y + 12z = 33
4x + 6y – 2z = 10
13x + 10z = 43
multiplying equation 1 × 3
multiplying equation 2 × 2
equation 4 by adding the equations
Now eliminate y from equation 2 and equation 3
8x + 12y – 4z = 20
3x + 12y – 6z = 9
5x + 2z = 11
multiplying equation 1 × 4
multiplying equation 3 × 3
equation 5 by subtracting the equations
Now equation 4 and equation 5 have 2 variables and the question is solved like a system
of 2 equations.
Remember, to get to this step you could eliminate any variable, not just y.
15
Module A50 − Systems of Equations
13x + 10z = 43
5x + 2z = 11
equation 4
equation 5
Now let’s eliminate the z’s,
13x + 10z = 43
25x + 10z = 55
−12x = −12
equation 4 doesn’t need to multiplied.
multiplying equation 4 × 5
by subtracting the equations
x=1
Substitute x = 1 into equation 4. We could substitute into either equation 4 or equation 5.
13(1) + 10z = 43
10z = 30
z=3
Substitute x = 1, z = 3 into equation 1. We could use equations 1, 2 or 3.
3x – 2y + 4z = 11
3(1) – 2y + 4(3) = 11
3 – 2y + 12 = 11
−2y + 15 = 11
−2y = −4
y=2
Solution is (1,2,3)
Check the solution.
3x – 2y + 4z =
11
2x + 3y – z =
5
x + 4y – 2z =
= 3(1) – 2(2) + 4(3)
= 2(1) + 3(2) – (3)
= (1) + 4(2) – 2(3)
= 3 – 4 + 12
=2+6−3
=1+8−6
= 11
=5
=3
checks
checks
checks
16
Module A50 − Systems of Equations
3
Experiential Activity Four
Solve the following systems of 3 equations.
1. 2x + 3y + z = 15
3x + 2y – z = 10
4x + y + 2z = 15
2. 4a – 2b + 3c = 27
2a + 3b – 4c = −6
3a + 5b – 2c = 12
3. 3a – 2b – 3c = 22
2a – 3b + 4c = 0
4a – b – 2c = 16
4. 2x + 5y – 3z = −7
3x – 2y – 4z = 16
5x + 2y – 5z = 4
Show Me.
5. 3a + 4b – 2c = −5
5a + 7b + 6c = 1
2a – 13b + 5c = 3
Experiential Activity Four Answers
Check your answers by substituting the variable solutions into the original equations.
The solutions should work for all equations in the system.
1. x = 2
y=3
z=2
2. x = 4
y=2
z=5
3. x = 1.3
y = −3.8
z = −3.5
4. x = −2
y = −3
z = −4
5. x = −1
y=0
z=1
17
Module A50 − Systems of Equations
OBJECTIVE FIVE
When you complete this objective you will be able to…
Solve word problems requiring the solution of 2 by 2 systems of equations.
Exploration Activity
EXAMPLE 1
A given fraction is such that if 2 is added to the numerator the fraction is equivalent to
7
. If 2 is added to the denominator of the original fraction the fraction is equivalent to
8
1
a
. Find the original fraction in reduced form .
2
b
SOLUTION:
Let the numerator of the original fraction = x.
Let the denominator of the original fraction = y; thus the fraction is:
x
y
then
2+ x 7
=
equation 1.
y
8
Here 2 is added to the numerator, and the fraction =
and
x
1
=
equation 2
2+ y 2
here 2 is added to the denominator, and the fraction is
Write equations 1 and 2 in standard form and solve.
8x – 7y = −16
2x – y = 2
equation 3 Å equation 1 written in standard form
equation4 Å equation 2 written in standard form
Now make the x coefficients the same and subtract
8x – 7y = −16
8x – 4y = 8
−3y = −24
equation 3
equation 4
equation 3 doesn’t need to multiplied.
equation 4 × 4
y=8
18
Module A50 − Systems of Equations
7
8
1
2
Substitute y = 8 into equation 1
2+ x 7
= Å Solve for x
8
8
2+x=7
x=5
THUS the ORIGINAL FRACTION:
x 5
=
y 8
Check the solution.
Is
2+5 7
= ?
8
8
Yes.
Is
5
1
= ?
2+8 2
Yes.
19
Module A50 − Systems of Equations
Experiential Activity Five
1. A contractor paid 13 people a total of $3965.25 for a week’s work. Some of the
workers were carpenters who were paid $340.25 per week and some were helpers
who were paid $225.75 per week. Determine the number of carpenters who were
paid.
2. A trucker drove at an average speed of 80 km/h from Calgary to a lumber mill in
B.C. Poor weather conditions resulted in the return trip averaging 60 km/h.
Determine the driving time for each direction if the total driving time is 7 hours.
Also determine the distance between Calgary and the lumber mill.
3. Two anglers took 9 h to canoe 45 km upstream to their favorite fishing hole in Cape
Breton National Park. The return trip took 5 h. Determine the speed of the stream’s
current in km/h.
4. A laboratory technician wished to make 150 L of 55% alcohol solution by mixing
45% alcohol and 70% alcohol solutions. Determine the number of litres of each
solution that should be used. Show Me.
5. A 7 kg mixture of cashews and pecans was sold for $32.80. If cashews and pecans
sell for $4.40/kg and $4.90/kg respectively, determine the number of kilograms of
cashews in the mixture.
Experiential Activity Five Answers
1. 9 carpenters
2. Time to drive from Calgary to the mill: 3 hours
Time to drive from Calgary to the mill: 4 hours
Distance between Calgary and the mill: 240 km
3. 90 L of 45% solution
60 L of 70% solution
4. 3 kg of cashews
20
Module A50 − Systems of Equations
OBJECTIVE SIX
When you complete this objective you will be able to…
Solve word problems requiring the solution of 3 by 3 systems of equations.
Exploration Activity
EXAMPLE 1
An electrical circuit contains three resistors. The sum of the currents in the resistors is 20
amperes. The first current minus the other two currents is 0 amperes. Twice the first
current plus the second current plus three times the third current is 38 amperes.
Determine the three currents.
SOLUTION:
Let the currents be A, B and C.
Now we set up the equations by translating each English statement from the question into
an algebraic statement.
1st statement:
2nd statement
3rd statement
A + B + C = 20
A–B–C=0
2A + B + 3C = 38
equation 1
equation 2
equation 3
Let’s eliminate the B’s first.
Add equations 1 and 2
A + B + C = 20
A–B–C=0
2A = 20
A = 10
equation 1
equation 2
equation 4
Here we are lucky because the C’s also disappeared and we find A right away to be 10.
Now add equations 2 and 3
A–B–C=0
2A + B + 3C = 38
3A + 2C = 38
equation 2
equation 3
equation 5
We can substitute A = 10 into equation 5 to get C
3(10)+ 2C = 38
C=4
21
Module A50 − Systems of Equations
Now substitute A = 10 and C = 4 into equation 1 to find B
(10) + B + (4) = 20
so B = 6
Check in equation 2
A–B–C=0
10 – 6 – 4 = 0 The values work.
The 3 currents are 10 amperes, 6 amperes, and 4 amperes.
Would you like more information? Visit the following web site for more practice with
systems of equations.
http://library.thinkquest.org/20991/alg2/systems.html?tqskip=1
22
Module A50 − Systems of Equations
Experiential Activity Six
Solve the following problems algebraically using systems of equations.
1. A person has incurred three financial debts that total $549.00. The first debt
minus 7 times the second debt minus the third debt is $ −909.00. Then 2 times
the first debt plus the sum of the other two debts is $702.00. Determine each of
the three debts.
2. Three machinists can produce a total of 50 drill bits. The first machinist’s
production plus 2 times the second machinist’s production plus the third
machinist’s production is 73 bits. Then 9 times the first machinist’s production
plus the other two machinist’s production is 146 bits. Determine the number of
bits each machinist can produce.
3. The sum of three numbers is –8. The first number plus 7 times the second
number plus the third number is –92. Then 2 times the first plus the other two is
–15. Find the numbers. Show Me.
Experiential Activity Six Answers
1. First debt = $153.00
Second debt = $111.00
Third debt = $285.00
2. First machinist’s production = 12
Second machinist’s production = 23
Third machinist’s production = 15
3. First number = –7
Second number = −14
Third number = 13
Practical Application Activity
Complete the Systems of Equations assignment in TLM.
Summary
This module presented the student with some of the basic skills necessary to solve 2 × 2
and 3 × 3 systems of linear equations.
The student should realize that these skills are a very significant part of any Algebra
course and are important for use in future mathematics courses.
23
Module A50 − Systems of Equations
Appendix
MODULE 50 GRAPHS
1.
2.
y
-15
-10
y
15
15
10
10
5
5
-5
5
10
-15
15
-10
-5
5
10
15
x
x
-5
-5
-10
-10
-15
-15
3.
4.
y
y
-15
-10
15
15
10
10
5
5
-5
5
10
-15
15
-10
-5
5
x
10
15
x
-5
-5
-10
-10
-15
-15
5.
6.
y
-15
-10
y
15
15
10
10
5
5
-5
5
10
-15
15
-10
-5
5
10
15
x
x
-5
-5
-10
-10
-15
-15
7.
8.
y
-15
-10
y
15
15
10
10
5
5
-5
5
10
-15
15
-10
-5
5
x
10
15
x
-5
-5
-10
-10
-15
-15
9.
10.
y
-15
-10
y
15
15
10
10
5
5
-5
5
10
-15
15
-10
-5
5
10
15
x
x
-5
-5
-10
-10
-15
-15
11.
12.
y
-15
-10
y
15
15
10
10
5
5
-5
5
10
-15
15
-10
-5
5
x
10
15
x
-5
-5
-10
-10
-15
-15
13.
14.
y
-15
-10
y
15
15
10
10
5
5
-5
5
10
-15
15
-10
-5
5
10
15
x
x
-5
-5
-10
-10
-15
-15
15.
16.
y
-15
-10
y
15
15
10
10
5
5
-5
5
10
-15
15
-10
-5
5
x
10
15
x
-5
-5
-10
-10
-15
-15
17.
18.
y
-15
-10
y
15
15
10
10
5
5
-5
5
10
-15
15
-10
-5
5
10
15
x
x
-5
-5
-10
-10
-15
-15
19.
20.
y
-15
-10
y
15
15
10
10
5
5
-5
5
10
-15
15
-10
-5
5
x
10
15
x
-5
-5
-10
-10
-15
-15