OpenStax-CNX module: m21986
1
Systems of Linear Equations:
Elimination by Addition
∗
Wade Ellis
Denny Burzynski
This work is produced by OpenStax-CNX and licensed under the
Creative Commons Attribution License 3.0†
Abstract
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Beginning with
the graphical solution of systems, this chapter includes an interpretation of independent, inconsistent,
and dependent systems and examples to illustrate the applications for these systems. The substitution
method and the addition method of solving a system by elimination are explained, noting when to use
each method. The ve-step method is again used to illustrate the solutions of value and rate problems
(coin and mixture problems), using drawings that correspond to the actual situation. Objectives of this
module: know the properties used in the addition method, be able to use the addition method to solve
a system of linear equations, know what to expect when using the addition method with a system that
consists of parallel or coincident lines.
1 Overview
• The Properties Used in the Addition Method
• The Addition Method
• Addition and Parallel or Coincident Lines
2 The Properties Used in the Addition Method
Another method of solving a system of two linear equations in two variables is called the method of
elimination by addition. It is similar to the method of elimination by substitution in that the process
eliminates one equation and one variable. The method of elimination by addition makes use of the following
two properties.
1. If A, B , and C are algebraic expressions such that
A =
B
C
D
A+C
∗ Version
=
= B+D
and
then
1.4: Jun 1, 2009 12:40 pm -0500
† http://creativecommons.org/licenses/by/3.0/
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OpenStax-CNX module: m21986
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2. ax + (−ax) = 0
Property 1 states that if we add the left sides of two equations together and the right sides of the same two
equations together, the resulting sums will be equal. We call this adding equations. Property 2 states
that the sum of two opposites is zero.
3 The Addition Method
To solve a system of two linear equations in two variables by addition,
1. Write, if necessary, both equations in general form, ax + by = c.
2. If necessary, multiply one or both equations by factors that will produce opposite coecients for one
of the variables.
3. Add the equations to eliminate one equation and one variable.
4. Solve the equation obtained in step 3.
5. Do one of the following:
(a) Substitute the value obtained in step 4 into either of the original equations and solve to obtain the
value of the other variable,
or
(b) Repeat steps 1-5 for the other variable.
6. Check the solutions in both equations.
7. Write the solution as an ordered pair.
The addition method works well when the coecient of one of the variables is 1 or a number other than 1.
4 Sample Set A
Example 1
Solve {
x−y =2
(1)
3x + y = 14
(2)
Step 1: Both equations appear in the proper form.
Step 2: The coecients of y are already opposites, 1 and −1, so there is no need for a multiplication.
Step 3: Add the equations.
x−y =2
3x + y = 14
4x+0=16
Step 4: Solve the equation 4x = 16.
4x = 16
x=4
The problem is not solved yet; we still need the value of y .
Step 5: Substitute x = 4 into either of the original equations. We will use equation 1.
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OpenStax-CNX module: m21986
4−y
−y
y
=
3
Solve for y.
2
= −2
=
2
We now have x = 4, y = 2.
Step 6: Substitute x = 4 and y = 2 into both the original equations for a check.
(1)
x−y
=
2
4−2
=
2
2
=
(2)
3x + y
=
14
Is this correct?
3 (4) + 2
=
14
Is this correct?
2 Yes, this is correct.
12 + 2
=
14
Is this correct?
14
=
14
Yes, this is correct.
Step 7: The solution is (4, 2) .
The two lines of this system intersect at (4, 2) .
5 Practice Set A
Solve each system by addition.
Exercise 1
{
(Solution on p. 11.)
x+y =6
2x − y = 0
Exercise 2
{
(Solution on p. 11.)
x + 6y = 8
−x − 2y = 0
6 Sample Set B
Solve the following systems using the addition method.
Example 2
Solve {
6a − 5b = 14
(1)
2a + 2b = −10
(2)
Step 1: The equations are already in the proper form, ax + by = c.
Step 2: If we multiply equation (2) by 3, the coecients of a will be opposites and become 0
upon addition, thus eliminating a.
{
6a − 5b = 14
−3 (2a + 2b) = −3 (10)
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→
{
6a − 5b = 14
−6a − 6b = 30
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Step 3: Add the equations.
6a − 5b = 14
−6a − 6b = 30
0−11b=44
Step 4: Solve the equation −11b = 44.
−11b = 44
b = −4
Step 5: Substituteb = −4 into either of the original equations. We will use equation 2.
2a + 2b = −10
2a + 2 (−4)
2a − 8
Solve for a.
−10
=
= −10
2a
=
−2
a
=
−1
We now have a = −1 and b = −4.
Step 6: Substitute a = −1 and b = −4 into both the original equations for a check.
6a − 5b =
(1)
14
(2)
2a + 2b = −10
6 (−1) − 5 (−4)
=
14
Is this correct?
2 (−1) + 2 (−4)
= −10
Is this correct?
−6 + 20
=
14
Is this correct?
−2 − 8
= −10
Is this correct?
14
=
14
Yes, this is correct.
−10
= −10 Yes, this is correct.
Step 7: The solution is (−1, −4) .
Example 3
Solve {
3x + 2y = −4
(1)
4x = 5y + 10
(2)
Step 1: Rewrite the system in the proper form.
{
3x + 2y = −4
(1)
4x − 5y = 10
(2)
Step 2: Since the coecients of y already have opposite signs, we will eliminate y .
Multiply equation (1) by 5, the coecient of y in equation 2.
Multiply equation (2) by 2, the coecient of y in equation 1.
{
5 (3x + 2y) = 5 (−4)
2 (4x − 5y) = 2 (10)
Step 3: Add the equations.
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→
{
15x + 10y = −20
8x − 10y = 20
OpenStax-CNX module: m21986
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15x + 10y = −20
8x − 10y = 20
23x+0=0
Step 4: Solve the equation 23x = 0
23x = 0
x=0
Step 5: Substitute x = 0 into either of the original equations. We will use equation 1.
3x + 2y
= −4
3 (0) + 2y
= −4
0 + 2y
= −4
y
= −2
Solve for y.
We now have x = 0 and y = −2.
Step 6: Substitution will show that these values check.
Step 7: The solution is (0, −2) .
7 Practice Set B
Solve each of the following systems using the addition method.
Exercise 3
{
(Solution on p. 11.)
3x + y = 1
5x + y = 3
Exercise 4
{
x + 4y = 1
x − 2y = −5
Exercise 5
{
−x + 2y = −2
(Solution on p. 11.)
5x − 3y = 1
8x − 6y = 4
Exercise 7
{
(Solution on p. 11.)
2x + 3y = −10
Exercise 6
{
(Solution on p. 11.)
3x − 5y = 9
4x + 8y = 12
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(Solution on p. 11.)
OpenStax-CNX module: m21986
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8 Addition And Parallel Or Coincident Lines
When the lines of a system are parallel or coincident, the method of elimination produces results identical
to that of the method of elimination by substitution.
Addition and Parallel Lines
If computations eliminate all variables and produce a contradiction, the two lines of the system are parallel
and the system is called inconsistent.
Addition and Coincident Lines
If computations eliminate all variables and produce an identity, the two lines of the system are coincident
and the system is called dependent.
9 Sample Set C
Example 4
Solve {
2x − y = 1
(1)
4x − 2y = 4
(2)
Step 1: The equations are in the proper form.
Step 2: We can eliminate x by multiplying equation (1) by 2.
{
−2 (2x − y) = −2 (1)
→
4x − 2y = 4
{
−4x + 2y = −2
4x − 2y = 4
Step 3: Add the equations.
−4x + 2y = −2
4x − 2y = 4
0+0=2
0=2
This is false and is therefore a contradiction. The lines of this system are parallel. This system is
inconsistent.
Example 5
Solve {
4x + 8y = 8
(1)
3x + 6y = 6
(2)
Step 1: The equations are in the proper form.
Step 2: We can eliminate x by multiplying equation (1) by 3 and equation (2) by 4.
{
−3 (4x + 8y) = −3 (8)
4 (3x + 6y) = 4 (6)
Step 3: Add the equations.
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→
{
−12x − 24y = −24
12x + 24y = 24
OpenStax-CNX module: m21986
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−12x − 24y = −24
12x + 24y = 24
0+0=0
0=0
This is true and is an identity. The lines of this system are coincident.
This system is dependent.
10 Practice Set C
Solve each of the following systems using the addition method.
Exercise 8
{
−x + 2y = 6
−6x + 12y = 1
Exercise 9
{
(Solution on p. 11.)
(Solution on p. 11.)
4x − 28y = −4
x − 7y = −1
11 Exercises
For the following problems, solve the systems using elimination by addition.
Exercise 10
{
(Solution on p. 11.)
x + y = 11
x − y = −1
Exercise 11
{
x + 3y = 13
x − 3y = −11
Exercise 12
{
(Solution on p. 11.)
3x − 5y = −4
−4x + 5y = 2
Exercise 13
{
2x − 7y = 1
5x + 7y = −22
Exercise 14
{
−3x + 4y = −24
3x − 7y = 42
Exercise 15
{
8x + 5y = 3
9x − 5y = −71
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(Solution on p. 11.)
OpenStax-CNX module: m21986
Exercise 16
{
8
(Solution on p. 11.)
−x + 2y = −6
x + 3y = −4
Exercise 17
{
4x + y = 0
3x + y = 0
Exercise 18
{
(Solution on p. 11.)
x + y = −4
−x − y = 4
Exercise 19
{
−2x − 3y = −6
2x + 3y = 6
Exercise 20
{
(Solution on p. 11.)
3x + 4y = 7
x + 5y = 6
Exercise 21
{
4x − 2y = 2
7x + 4y = 26
Exercise 22
{
(Solution on p. 11.)
3x + y = −4
5x − 2y = −14
Exercise 23
{
5x − 3y = 20
−x + 6y = −4
Exercise 24
{
(Solution on p. 11.)
6x + 2y = −18
−x + 5y = 19
Exercise 25
{
x − 11y = 17
2x − 22y = 4
Exercise 26
{
(Solution on p. 11.)
−2x + 3y = 20
−3x + 2y = 15
Exercise 27
{
−5x + 2y = −4
−3x − 5y = 10
Exercise 28
{
−3x − 4y = 2
−9x − 12y = 6
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(Solution on p. 11.)
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Exercise 29
{
3x − 5y = 28
−4x − 2y = −20
Exercise 30
{
(Solution on p. 11.)
6x − 3y = 3
10x − 7y = 3
Exercise 31
{
−4x + 12y = 0
−8x + 16y = 0
Exercise 32
{
(Solution on p. 11.)
3x + y = −1
12x + 4y = 6
Exercise 33
{
8x + 5y = −23
−3x − 3y = 12
Exercise 34
{
(Solution on p. 11.)
2x + 8y = 10
3x + 12y = 15
Exercise 35
{
4x + 6y = 8
6x + 8y = 12
Exercise 36
{
(Solution on p. 11.)
10x + 2y = 2
−15x − 3y = 3
Exercise 37
{
x + 34 y = − 12
3
5x
+ y = − 75
Exercise 38
{
(Solution on p. 11.)
x + 31 y = 43
−x + 61 y = 23
Exercise 39
{
8x − 3y = 25
4x − 5y = −5
Exercise 40
{
−10x − 4y = 72
9x + 5y = 39
Exercise 41
{
12x + 16y = −36
−10x + 12y = 30
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(Solution on p. 11.)
OpenStax-CNX module: m21986
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Exercise 42
{
(Solution on p. 12.)
25x − 32y = 14
−50x + 64y = −28
12 Exercises For Review
Exercise 43
5
−5
( here1 ) Simplify and write 2x−3 y 4
2xy −6
so that only positive exponents appear.
Exercise 44
√
√
( here2 ) Simplify 8 + 3 50.
(Solution on p. 12.)
Exercise 45
√
( here3 ) Solve the radical equation 2x + 3 + 5 = 8.
Exercise 46
( here4 ) Solve by graphing {
(Solution on p. 12.)
x+y =4
3x − y = 0
Exercise 47
( here5 ) Solve using the substitution method: {
3x − 4y = −11
5x + y = −3
1 "Basic Operations with Real Numbers: Multiplication and Division of Signed Numbers"
<http://cnx.org/content/m21872/latest/>
2 "Roots, Radicals, and Square Root Equations: Addition and Subtraction of Square Root Expressions"
<http://cnx.org/content/m21957/latest/>
3 "Roots, Radicals, and Square Root Equations: Square Root Equations with Applications"
<http://cnx.org/content/m21965/latest/>
4 "Systems of Linear Equations: Solutions by Graphing" <http://cnx.org/content/m18881/latest/>
5 "Systems of Linear Equations: Elimination by Substitution" <http://cnx.org/content/m21984/latest/>
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OpenStax-CNX module: m21986
Solutions to Exercises in this Module
Solution to Exercise (p. 3)
(2, 4)
Solution to Exercise (p. 3)
(−4, 2)
Solution to Exercise (p. 5)
(1, −2)
Solution to Exercise (p. 5)
(−3, 1)
Solution to Exercise (p. 5)
(−2, −2)
Solution to Exercise (p. 5)
(−1, −2)
Solution to Exercise (p. 5)
(3, 0)
Solution to Exercise (p. 7)
inconsistent
Solution to Exercise (p. 7)
dependent
Solution to Exercise (p. 7)
(5, 6)
Solution to Exercise (p. 7)
(2, 2)
Solution to Exercise (p. 7)
(0, −6)
Solution to Exercise (p. 7)
(2, −2)
Solution to Exercise (p. 8)
dependent
Solution to Exercise (p. 8)
(1, 1)
Solution to Exercise (p. 8)
(−2, 2)
Solution to Exercise (p. 8)
(−4, 3)
Solution to Exercise (p. 8)
(−1, 6)
Solution to Exercise (p. 8)
dependent
Solution to Exercise (p. 9)
(1, 1)
Solution to Exercise (p. 9)
inconsistent
Solution to Exercise (p. 9)
dependent
Solution to Exercise (p. 9)
inconsistent
Solution to Exercise (p. 9)
(0, 4)
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OpenStax-CNX module: m21986
Solution to
Exercise (p. 9)
519
− 258
7 , 7
Solution to Exercise (p. 9)
dependent
Solution
to Exercise (p. 10)
√
17 2
Solution to Exercise (p. 10)
(1, 3)
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12