273
4.3. SOLVING SYSTEMS BY ELIMINATION
4.3
Solving Systems by Elimination
When both equations of a system are in standard form Ax + By = C, then
a process called elimination is usually the best procedure to use to find the
solution of the system. Elimination is based on two simple ideas, the first of
which should be familiar.
1. Multiplying both sides of an equation by a non-zero number does not
change its solutions. Thus, the equation
x + 3y = 7
(4.19)
will have the same solutions (it’s the same line) as the equation obtained
by multiplying equation (4.19) by 2.
2x + 6y = 14
(4.20)
2. Adding two true equations produces another true equation. For example,
consider what happens when you add 4 = 4 to 5 = 5.
4
5
9
=
=
=
4
5
9
Even more importantly, consider what happens when you add two equations that have (2, 1) as a solution. The result is a third equation whose
graph also passes through the solution.
x+y =3
x +
x −
2x
x
y
y
=
=
=
=
3
1
4
2
y
5
5
−5
x−y =1
−5
x
x=2
Fact. Adding a multiple of an equation to a second equation produces an
equation that passes through the same solution as the first two equations.
274
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
One more important thing to notice is the fact that when we added the
equations
x + y = 3
x − y = 1
2x
= 4
the variable y was eliminated. This is where the elimination method gets its
name. The strategy is to somehow add the equations of a system with the
intent of eliminating one of the unknown variables.
However, sometimes you need to do a little bit more than simply add the
equations. Let’s look at an example.
You Try It!
Solve the following system of
equations:
EXAMPLE 1. Solve the following system of equations.
x + 3y = 14
−8x − 3y = −28
x + 2y = −5
(4.21)
2x − y = −5
(4.22)
Solution: Our focus will be on eliminating the variable x. Note that if we
multiply equation (4.21) by −2, then add the result to equation (4.22), the x
terms will be eliminated.
y
5
2x − y = −5
(−3, −1)
−5
5
x + 2y = −5
Figure 4.38: x + 2y = −5
and 2x − y = −5 intersect at
(−3, −1).
Answer: (2, 4).
4y
y
=
=
10
−5
−
5y
=
5
Multiply equation (4.21) by −2.
Equation (4.22).
Add the equations.
Divide both sides of −5y = 5 by −5 to get y = −1.
To find the corresponding value of x, substitute −1 for y in equation (4.21)
(or equation (4.22)) and solve for x.
x
−5
−2x −
2x −
x + 2y = −5
x + 2(−1) = −5
x = −3
Equation (4.21)
Substitute −1 for y.
Solve for x.
Check: To check, we need to show that the point (x, y) = (−3, 1) satisfies
both equations.
Substitute (x, y) = (−3, −1) into
equation (4.21).
Substitute (x, y) = (−3, −1) into
equation (4.22).
x + 2y = −5
−3 + 2(−1) = −5
2x − y = −5
2(−3) − (−1) = −5
−5 = −5
−5 = −5
Thus, the point (x, y) = (−3, −1) satisfies both equations and is therefore the
solution of the system.
4.3. SOLVING SYSTEMS BY ELIMINATION
275
To show that you have the option of which variable you choose to eliminate,
let’s try Example 1 a second time, this time eliminating y instead of x.
You Try It!
EXAMPLE 2. Solve the following system of equations.
x + 2y = −5
(4.23)
2x − y = −5
(4.24)
y
Solution: This time we focus on eliminating the variable y. We note that if
we multiply equation (4.24) by 2, then add the result to equation (4.23), the y
terms will be eliminated.
x +
4x −
2y
2y
5x
= −5
= −10
Equation (4.23).
Multiply equation (4.24) by 2.
= −15
Add the equations.
x + 2y = −5
y = −1
2x − y = −5
x
−5
Divide both sides of 5x = −15 by 5 to get x = −3.
To find the corresponding value of y, substitute −3 for x in equation (4.23)
(or equation (4.24)) and solve for y.
−3 + 2y = −5
2y = −2
5
(−3, −1)
−5
5
x + 2y = −5
Figure 4.39: x + 2y = −5
and 2x − y = −5 intersect at
(−3, −1).
Equation (4.23)
Substitute −3 for x.
Add 3 to both sides.
Divide both sides by 2.
Hence, (x, y) = (−3, −1), just as in Example 1, is the solution of the system.
Sometimes elimination requires a thought process similar to that of finding
a common denominator.
You Try It!
EXAMPLE 3. Solve the following system of equations.
3x + 4y = 12
2x − 5y = 10
(4.25)
(4.26)
Solution: Let’s focus on eliminating the x-terms. Note that if we multiply
equation (4.25) by 2, then multiply equation (4.26) by −3, the x-terms will be
eliminated when we add the resulting equations.
6x +
−6x +
8y
15y
=
24
= −30
Multiply equation (4.25) by 2.
Multiply equation (4.26) by −3.
23y
=
Add the equations.
−6
Solve the following system of
equations:
−14x + 9y = 94
7x + 3y = −62
276
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
Hence, y = −6/23.
At this point, we could substitute y = −6/23 in either equation, then solve
the result for x. However, working with y = −6/23 is a bit daunting, particularly in the light of elimination being easier. So let’s use elimination again,
this time focusing on eliminating y. Note that if we multiply equation (4.25)
by 5, then multiply equation (4.26) by 4, when we add the results, the y-terms
will be eliminated.
y
5
x
8
2x − 5y = 10
=
=
60
40
23x
=
100
Multiply equation (4.25) by 5.
Multiply equation (4.26) by 4.
Add the equations.
Thus, x = 100/23, and the system of the system is (x, y) = (100/23, −6/23).
(100/23, −6/23)
−2
15x + 20y
8x − 20y
Check: Let’s use the graphing calculator to check the solution. First, store
100/23 in X, then −6/23 in Y (see Figure 4.41). Next, enter the left-hand sides
of equations (4.25) and (4.26).
3x + 4y = 12
−5
Figure 4.40: 3x + 4y = 12
and 2x − 5y = 10 intersect at
(100/23, −6/23).
Figure 4.41: Enter 100/23 in X,
−6/23 in Y.
Answer: (−8, −2).
Figure 4.42: Enter the left-hand
sides of equations (4.25) and (4.26).
Note that both calculations in Figure 4.42 provide the correct right-hand sides
for equations (4.25) and (4.26). Thus, the solution (x, y) = (100/23, −6/23)
checks.
Exceptional Cases
In the previous section, we saw that if the substitution method led to a false
statement, then we have parallel lines. The same thing can happen with the
elimination method of this section.
You Try It!
Solve the following system of
equations:
5x − 4y = −16
15x − 12y = 49
EXAMPLE 4. Solve the following system of equations.
x+y =3
(4.27)
2x + 2y = −6
(4.28)
277
4.3. SOLVING SYSTEMS BY ELIMINATION
Solution: Let’s focus on eliminating the x-terms. Note that if we multiply
equation (4.27) by −2, the x-terms will be eliminated when we add the resulting
equations.
−2x −
2x +
2y
2y
=
=
−6
−6
0
=
−12
Multiply equation (4.27) by −2.
Equation (4.28).
Add the equations.
Because of our experience with this solving this exceptional case with substitution, the fact that both variables have disappeared should not be completely
surprising. Note that this last statement is false, regardless of the values of x
and y. Hence, the system has no solution.
Indeed, if we find the intercepts of each equation and plot them, then we can
easily see that the lines of this system are parallel (see Figure 4.43). Parallel
lines never intersect, so the system has no solutions.
y
5
(0, 3)
(3, 0)
(−3, 0)
−5
5
x
x+y =3
(0, −3)
−5
2x + 2y = −6
Figure 4.43: The lines x + y = 3 and 2x + 2y = −6 are parallel.
Answer: no solution
In the previous section, we saw that if the substitution method led to a true
statement, then we have the same lines. The same thing can happen with the
elimination method of this section.
You Try It!
EXAMPLE 5. Solve the following system of equations.
x − 7y = 4
(4.29)
−3x + 21y = −12
(4.30)
Solve the following system of
equations:
2x − 7y = 4
8x − 28y = 16
278
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
y
5
(−3, −1)
(0, −4/7)
Solution: If we are not on automatic pilot late at night doing our homework,
we might recognize that the equations 4.29 and (4.30) are identical. But it’s
also conceivable that we don’t see that right away and begin the elimination
x − 7y = 4 method. Let’s multiply the first equation by 3, then add. This will eliminate
x the x-terms.
(4, 0)
−3x + 21y = −12
−5
Figure 4.44: x − 7y = 4 and
−3x + 21y = −12 are the
same line. Infinite number of
solutions.
Answer: There are an
infinite number of solutions.
Examples of solution points
are (2, 0), (9, 2), and
(−5, −2).
3x − 21y
−3x + 21y
=
=
12
−12
0 =
0
Multiply equation (4.29) by 3.
Equation (4.30).
Add the equations.
Again, all of the variables have disappeared! However, this time the last statement is true, regardless of the values of x and y.
Notice that if we multiply equation (4.29) by −3, then we have two identical
equations.
−3x + 21y
−3x + 21y
=
=
−12
−12
Multiply equation (4.29) by 3.
Equation (4.30).
The equations are identical! Hence, there are an infinite number of points of
intersection. Indeed, any point on either line is a solution. Example points of
solution are (−3, −1), (0, −4/7), and (4, 0).
279
4.3. SOLVING SYSTEMS BY ELIMINATION
❧ ❧ ❧
Exercises
❧ ❧ ❧
In Exercises 1-8, use the elimination method to solve each of the following systems. Check your result
manually, without the assistance of a calculator.
1.
x + 4y
9x − 7y
=
=
0
−43
2.
x + 6y
5x − 9y
=
=
−53
47
3.
6x + y
4x + 2y
=
=
8
0
4.
4x +
−2x +
y
6y
=
18
= −22
5.
−8x + y
4x + 3y
6.
2x +
7x +
7.
x + 8y
−5x − 9y
=
=
41
−50
8.
x − 4y
−2x − 6y
=
=
−31
−36
y
8y
=
=
−56
56
= 21
= 87
In Exercises 9-16, use the elimination method to solve each of the following systems.
9.
−12x + 9y
−6x − 4y
=
=
0
−34
13.
2x − 6y
−3x + 18y
=
=
28
−60
10.
−27x − 5y
−9x − 3y
=
=
148
60
14.
−8x − 6y
4x + 30y
=
=
96
−156
11.
27x −
−3x −
6y
5y
= −96
=
22
15.
−32x +
8x −
=
=
−238
64
12.
−8x +
2x −
8y
9y
= −32
=
15
16.
12x + 6y
−2x + 7y
7y
4y
=
=
30
51
In Exercises 17-24, use the elimination method to solve each of the following systems.
17.
3x −
−2x −
7y
2y
= −75
= −10
21.
−9x − 2y
5x − 3y
=
=
28
−32
18.
−8x +
−7x +
3y
8y
= 42
= 26
22.
−8x − 2y
6x + 3y
=
=
−12
12
19.
9x − 9y
2x − 6y
23.
−3x − 5y
7x + 7y
=
=
−34
56
24.
−9x − 9y
7x + 4y
=
=
9
8
20.
−4x −
−7x −
8y
3y
=
=
−63
−34
= −52
= −14
280
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
In Exercises 25-32, use the elimination method to solve each of the following systems. Use your
calculator to check your solutions.
25.
26.
2x − 7y
7x + 6y
= −2
=
3
−9x −
5x −
4y
3y
=
=
4
−1
27.
2x +
−5x +
3y
5y
=
=
−2
2
28.
−5x +
−4x −
8y
7y
=
=
−3
3
29.
30.
9x +
−7x −
4y
9y
= −4
=
3
−3x −
4x +
5y
6y
= −4
=
1
31.
2x +
3x −
32.
6x −
−4x −
2y
5y
=
=
9y
8y
4
3
= −2
=
4
In Exercises 33-40, use the elimination method to determine how many solutions each of the following
system of equations has.
33.
x +
−8x −
7y
56y
34.
−8x +
56x −
y
7y
35.
16x −
−8x +
16y
8y
36.
3x −
−6x +
3y
6y
= −32
= 256
=
=
−53
371
= −256
=
128
=
=
42
−84
❧ ❧ ❧
37.
x −
2x −
38.
4x + y
28x + 7y
= −13
= 189
39.
x +
−4x −
9y
5y
=
73
= −44
40.
6x +
−5x −
y
6y
=
31
= −62
Answers
❧ ❧ ❧
1. (−4, 1)
13. (8, −2)
3. (2, −4)
15. (7, −2)
5. (8, 8)
17. (−4, 9)
7. (1, 5)
19. (−2, 5)
9. (3, 4)
11. (−4, −2)
4y
8y
21. (−4, 4)
23. (3, 5)
=
=
−37
54
4.3. SOLVING SYSTEMS BY ELIMINATION
281
25. (9/61, 20/61)
33. Infinite number of solutions
27. (−16/25, −6/25)
35. Infinite number of solutions
29. (−24/53, 1/53)
37. No solutions
31. (13/8, 3/8)
39. One solution