1. Solving Systems by Elimination
5-3
Solving Systems by Elimination
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Algebra 1
Holt Algebra 1

2. Warm Up
Simplify each expression.
1. 3x + 2y – 5x – 2y
2. 5(x – y) + 2x + 5y
3. 4y + 6x – 3(y + 2x)
4. 2y – 4x – 2(4y – 2x)
–2x 7x y
–6y
Write the least common multiple. 5.
3 and 6 6 6.
4 and 10 20 7.
6 and 8 24 8.
2 and 5 10

3. Objectives
Solve systems of linear equations in two variables by elimination.
Compare and choose an appropriate method for solving systems of linear equations.

4. Another method for solving systems of equations is elimination
Another method for solving systems of equations is elimination. Like substitution, the goal of elimination is to get one equation that has only one variable.
Remember that an equation stays balanced if you add equal amounts to both sides.
Consider the system Since
5x + 2y = 1, you can add 5x + 2y to one side of the first equation and 1 to the other side and the balance is maintained.

5. Since –2y and 2y have opposite coefficients, you can eliminate the y by adding the two equations. The result is one equation that has only one variable: 6x = –18.
When you use the elimination method to solve a system of linear equations, align all like terms in the equations. Then determine whether any like terms can be eliminated because they have opposite coefficients.

6. Solving Systems of Equations by Elimination
Step 1
Write the system so that like terms are aligned.
Step 2
Eliminate one of the variables and solve for the other variable.
Step 3
Substitute the value of the variable into one of the original equations and solve for the other variable.
Write the answers from Steps 2 and 3 as an ordered pair, (x, y), and check.
Step 4

7. Later in this lesson you will learn how to multiply one or more equations by a number in order to produce opposites that can be eliminated.

8. Example 1: Elimination Using Addition
3x – 4y = 10
Solve by elimination.
x + 4y = –2
Step 1
3x – 4y = 10
Align like terms. −4y and +4y are opposites.
x + 4y = –2
Step 2
Add the equations to eliminate y.
4x = 8
4x = 8
Simplify and solve for x.
4x = 8
x = 2
Divide both sides by 4.

9. Example 1 Continued
Step 3
x + 4y = –2
Write one of the original equations.
2 + 4y = –2
Substitute 2 for x.
– –2
4y = –4
Subtract 2 from both sides.
4y –4
y = –1
Divide both sides by 4.
Step 4 (2, –1)
Write the solution as an ordered pair.

10. Check It Out! Example 1
y + 3x = –2
Solve by elimination.
2y – 3x = 14
Align like terms. 3x and −3x are opposites.
Step 1
2y – 3x = 14
y + 3x = –2
Add the equations to eliminate x.
Step 2
3y = 12
3y = 12
Simplify and solve for y.
y = 4
Divide both sides by 3.

11. Check It Out! Example 1 Continued
Write one of the original equations.
Step 3 y + 3x = –2
4 + 3x = –2
Substitute 4 for y.
– –4
3x = –6
Subtract 4 from both sides.
Divide both sides by 3.
3x = –6
x = –2
Write the solution as an ordered pair.
Step 4
(–2, 4)

12. When two equations each contain the same term, you can subtract one equation from the other to solve the system. To subtract an equation, add the opposite of each term.

13. Example 2: Elimination Using Subtraction
2x + y = –5
Solve by elimination.
2x – 5y = 13
2x + y = –5
Step 1
Both equations contain 2x. Add the opposite of each term in the second equation.
–(2x – 5y = 13)
2x + y = –5
–2x + 5y = –13
0 + 6y = –18
Step 2
Eliminate x.
6y = –18
y = –3
Simplify and solve for y.

14. Example 2 Continued
Write one of the original equations.
Step 3
2x + y = –5
2x + (–3) = –5
Substitute –3 for y.
2x – 3 = –5
Add 3 to both sides.
2x = –2
Simplify and solve for x.
x = –1
Step 4
(–1, –3)
Write the solution as an ordered pair.

15. Remember to check by substituting your answer into both original equations.

16. Check It Out! Example 2
3x + 3y = 15
Solve by elimination.
–2x + 3y = –5
3x + 3y = 15
–(–2x + 3y = –5)
Step 1
Both equations contain 3y. Add the opposite of each term in the second equation.
3x + 3y = 15
+ 2x – 3y = +5
Step 2
5x = 20
Eliminate y.
5x = 20
x = 4
Simplify and solve for x.

17. Check It Out! Example 2 Continued
Write one of the original equations.
Step 3
3x + 3y = 15
3(4) + 3y = 15
Substitute 4 for x.
12 + 3y = 15
– –12
3y = 3
Subtract 12 from both sides.
Simplify and solve for y.
y = 1
Write the solution as an ordered pair.
(4, 1)
Step 4

18. In some cases, you will first need to multiply one or both of the equations by a number so that one variable has opposite coefficients.

19. Example 3A: Elimination Using Multiplication First
Solve the system by elimination.
x + 2y = 11
–3x + y = –5
Multiply each term in the second equation by –2 to get opposite y-coefficients.
x + 2y = 11
Step 1
–2(–3x + y = –5)
x + 2y = 11
+(6x –2y = +10)
Add the new equation to the first equation to eliminate y.
7x = 21
Step 2
7x = 21
x = 3
Solve for x.

20. Example 3A Continued
Write one of the original equations.
Step 3
x + 2y = 11
3 + 2y = 11
Substitute 3 for x.
– –3
2y = 8
Subtract 3 from both sides.
Solve for y.
y = 4
Step 4
(3, 4)
Write the solution as an ordered pair.

21. Example 3B: Elimination Using Multiplication First
Solve the system by elimination.
–5x + 2y = 32
2x + 3y = 10
Multiply the first equation by 2 and the second equation by 5 to get opposite x-coefficients
Step 1
2(–5x + 2y = 32)
5(2x + 3y = 10)
–10x + 4y = 64
+(10x + 15y = 50)
Add the new equations to
eliminate x.
Step 2
19y = 114
y = 6
Solve for y.

22. Example 3B Continued
Write one of the original equations.
Step 3
2x + 3y = 10
2x + 3(6) = 10
Substitute 6 for y.
2x + 18 = 10
–18 –18
2x = –8
Subtract 18 from both sides.
x = –4
Solve for x.
Step 4
Write the solution as an ordered pair.
(–4, 6)

23. Check It Out! Example 3a
Solve the system by elimination.
3x + 2y = 6
–x + y = –2
Multiply each term in the second equation by 3 to get opposite x-coefficients.
Step 1
3x + 2y = 6
3(–x + y = –2)
3x + 2y = 6
+(–3x + 3y = –6)
y = 0
Add the new equation to the first equation.
Simplify and solve for y.
5y = 0
y = 0
Step 2

24. Check It Out! Example 3a Continued
Write one of the original equations.
Step 3
–x + y = –2
–x + 3(0) = –2
Substitute 0 for y.
–x + 0 = –2
Solve for x.
–x = –2
x = 2
Step 4
Write the solution as an ordered pair.
(2, 0)

25. Check It Out! Example 3b
Solve the system by elimination.
2x + 5y = 26
–3x – 4y = –25
Multiply the first equation by 3 and the second equation by 2 to get opposite x-coefficients
Step 1
3(2x + 5y = 26)
+(2)(–3x – 4y = –25)
6x + 15y = 78
+(–6x – 8y = –50)
Add the new equations to
eliminate x.
y = 28
Step 2
y = 4
Solve for y.

26. Check It Out! Example 3b Continued
Write one of the original equations.
Step 3
2x + 5y = 26
2x + 5(4) = 26
Substitute 4 for y.
2x + 20 = 26
–20 –20
2X = 6
Subtract 20 from both sides.
Solve for x.
x = 3
Step 4
Write the solution as an ordered pair.
(3, 4)

27. Example 4: Application
Paige has $7.75 to buy 12 sheets of felt and card stock for her scrapbook. The felt costs $0.50 per sheet, and the card stock costs $0.75 per sheet. How many sheets of each can Paige buy?
Write a system. Use f for the number of felt sheets and c for the number of card stock sheets.
0.50f c = 7.75
The cost of felt and card stock totals $7.75.
f + c = 12
The total number of sheets is 12.

28. Example 4 Continued
Step 1
0.50f c = 7.75
Multiply the second equation by –0.50 to get opposite f-coefficients.
+ (–0.50)(f + c) = 12
0.50f c = 7.75
Add this equation to the first equation to eliminate f.
+ (–0.50f – 0.50c = –6)
0.25c = 1.75
Step 2
c = 7
Solve for c.
Write one of the original equations.
Step 3
f + c = 12
f + 7 = 12
Substitute 7 for c.
–7 –7
f = 5
Subtract 7 from both sides.

29. Example 4 Continued
Step 4
(7, 5)
Write the solution as an ordered pair.
Paige can buy 7 sheets of card stock and 5 sheets of felt.

30. Check It Out! Example 4
What if…? Sally spent $14.85 to buy 13 flowers. She bought lilies, which cost $1.25 each, and tulips, which cost $0.90 each. How many of each flower did Sally buy?
Write a system. Use l for the number of lilies and t for the number of tulips.
1.25l t = 14.85
The cost of lilies and tulips totals $14.85.
l + t = 13
The total number of flowers is 13.

31. Check It Out! Example 4 Continued
Step 1
1.25l + .90t = 14.85
Multiply the second equation by –0.90 to get opposite t-coefficients.
+ (–.90)(l + t) = 13
1.25l t = 14.85
+ (–0.90l – 0.90t = –11.70)
Add this equation to the first equation to eliminate t.
0.35l = 3.15
Step 2
Solve for l.
l = 9

32. Check It Out! Example 4 Continued
Step 3
Write one of the original equations.
l + t = 13
9 + t = 13
Substitute 9 for l.
– –9
Subtract 9 from both sides.
t = 4
Step 4
(9, 4)
Write the solution as an ordered pair.
Sally bought 9 lilies and 4 tulips.

33. All systems can be solved in more than one way
All systems can be solved in more than one way. For some systems, some methods may be better than others.

35. Lesson Quiz
Solve each system by elimination. 1. 2. 3.
2x + y = 25
(11, 3)
3y = 2x – 13
–3x + 4y = –18
(2, –3)
x = –2y – 4
–2x + 3y = –15
(–3, –7)
3x + 2y = –23
4. Harlan has $44 to buy 7 pairs of socks. Athletic socks cost $5 per pair. Dress socks cost $8 per pair. How many pairs of each can Harlan buy?
4 pairs of athletic socks and 3 pairs of dress socks

36. Solving Special Systems
5-4
Solving Special Systems
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Algebra 1
Holt Algebra 1

37. infinitely many solutions
Warm Up
Solve each equation.
1. 2x + 3 = 2x + 4
2. 2(x + 1) = 2x + 2
3. Solve 2y – 6x = 10 for y
no solution
infinitely many solutions
y =3x + 5
Solve by using any method.
y = 3x + 2
x – y = 8 4. 5.
(1, 5)
(6, –2)
2x + y = 7
x + y = 4

38. Objectives Solve special systems of linear equations in two variables.
Classify systems of linear equations and determine the number of solutions.

39. Vocabulary inconsistent system consistent system independent system

40. In Lesson 6-1, you saw that when two lines intersect at a point, there is exactly one solution to the system. Systems with at least one solution are called consistent.
When the two lines in a system do not intersect they are parallel lines. There are no ordered pairs that satisfy both equations, so there is no solution. A system that has no solution is an inconsistent system.

41. Example 1: Systems with No Solution
Show that has no solution.
y = x – 4
–x + y = 3
Method 1 Compare slopes and y-intercepts.
y = x – y = 1x – 4
Write both equations in slope-intercept form.
–x + y = y = 1x + 3
The lines are parallel because they have the same slope and different y-intercepts.
This system has no solution.

42.  Example 1 Continued y = x – 4 Show that has no solution. –x + y = 3
Method 2 Solve the system algebraically. Use the substitution method because the first equation is solved for y.
–x + (x – 4) = 3
Substitute x – 4 for y in the second equation, and solve.
–4 = 3 
False.
This system has no solution.

43. Show that has no solution. y = x – 4
Example 1 Continued
Show that has no solution.
y = x – 4
–x + y = 3
Check Graph the system.
– x + y = 3
The lines appear are parallel.
y = x – 4

44. Check It Out! Example 1
Show that has no solution.
y = –2x + 5
2x + y = 1
Method 1 Compare slopes and y-intercepts.
y = –2x y = –2x + 5
2x + y = y = –2x + 1
Write both equations in slope-intercept form.
The lines are parallel because they have the same slope and different y-intercepts.
This system has no solution.

45. Check It Out! Example 1 Continued
Show that has no solution.
y = –2x + 5
2x + y = 1
Method 2 Solve the system algebraically. Use the substitution method because the first equation is solved for y.
2x + (–2x + 5) = 1
Substitute –2x + 5 for y in the second equation, and solve.
5 = 1 
False.
This system has no solution.

46. Check It Out! Example 1 Continued
Show that has no solution.
y = –2x + 5
2x + y = 1
Check Graph the system.
y = –2x + 5
y = – 2x + 1
The lines are parallel.

47. If two linear equations in a system have the same graph, the graphs are coincident lines, or the same line. There are infinitely many solutions of the system because every point on the line represents a solution of both equations.

48. Example 2A: Systems with Infinitely Many Solutions
Show that has infinitely
many solutions.
y = 3x + 2
3x – y + 2= 0
Method 1 Compare slopes and y-intercepts.
Write both equations in slope-intercept form. The lines have the same slope and the same y-intercept.
y = 3x y = 3x + 2
3x – y + 2= y = 3x + 2
If this system were graphed, the graphs would be the same line. There are infinitely many solutions.

49.  Example 2A Continued y = 3x + 2 Show that has infinitely
many solutions.
y = 3x + 2
3x – y + 2= 0
Method 2 Solve the system algebraically. Use the elimination method.
Write equations to line up like terms.
y = 3x y − 3x = 2
3x − y + 2= −y + 3x = −2
Add the equations.
0 = 0 
True. The equation is an identity.
There are infinitely many solutions.

50. Caution!
0 = 0 is a true statement. It does not mean the system has zero solutions or no solution.

51. Check It Out! Example 2
Show that has infinitely
many solutions.
y = x – 3
x – y – 3 = 0
Method 1 Compare slopes and y-intercepts.
Write both equations in slope-intercept form. The lines have the same slope and the same y-intercept.
y = x – y = 1x – 3
x – y – 3 = y = 1x – 3
If this system were graphed, the graphs would be the same line. There are infinitely many solutions.

52. Check It Out! Example 2 Continued
Show that has infinitely
many solutions.
y = x – 3
x – y – 3 = 0
Method 2 Solve the system algebraically. Use the elimination method.
Write equations to line up like terms.
y = x – y = x – 3
x – y – 3 = –y = –x + 3
Add the equations.
0 = 0 
True. The equation is an identity.
There are infinitely many solutions.

53. Consistent systems can either be independent or dependent.
An independent system has exactly one solution. The graph of an independent system consists of two intersecting lines.
A dependent system has infinitely many solutions. The graph of a dependent system consists of two coincident lines.

55. Example 3A: Classifying Systems of Linear Equations
Classify the system. Give the number of solutions.
3y = x + 3
Solve
x + y = 1
3y = x y = x + 1
Write both equations in slope-intercept form.
x + y = 1
y = x + 1
The lines have the same slope and the same y-intercepts. They are the same.
The system is consistent and dependent. It has infinitely many solutions.

56. Example 3B: Classifying Systems of Linear equations
Classify the system. Give the number of solutions.
x + y = 5
Solve
4 + y = –x
x + y = y = –1x + 5
Write both equations in slope-intercept form.
4 + y = –x y = –1x – 4
The lines have the same slope and different y-intercepts. They are parallel.
The system is inconsistent. It has no solutions.

57. Example 3C: Classifying Systems of Linear equations
Classify the system. Give the number of solutions.
y = 4(x + 1)
Solve
y – 3 = x
y = 4(x + 1) y = 4x + 4
Write both equations in slope-intercept form.
y – 3 = x y = 1x + 3
The lines have different slopes. They intersect.
The system is consistent and independent. It has one solution.

58. Check It Out! Example 3a
Classify the system. Give the number of solutions.
x + 2y = –4
Solve
–2(y + 2) = x
Write both equations in slope-intercept form.
y = x – 2
x + 2y = –4
–2(y + 2) = x
y = x – 2
The lines have the same slope and the same y-intercepts. They are the same.
The system is consistent and dependent. It has infinitely many solutions.

59. Check It Out! Example 3b
Classify the system. Give the number of solutions.
y = –2(x – 1)
Solve
y = –x + 3
Write both equations in slope-intercept form.
y = –2(x – 1)
y = –2x + 2
y = –x + 3
y = –1x + 3
The lines have different slopes. They intersect.
The system is consistent and independent. It has one solution.

60. Check It Out! Example 3c
Classify the system. Give the number of solutions.
2x – 3y = 6
Solve
y = x
Write both equations in slope-intercept form.
y = x
2x – 3y = 6
y = x – 2
The lines have the same slope and different y-intercepts. They are parallel.
The system is inconsistent. It has no solutions.

61. Example 4: Application
Jared and David both started a savings account in January. If the pattern of savings in the table continues, when will the amount in Jared’s account equal the amount in David’s account?
Use the table to write a system of linear equations. Let y represent the savings total and x represent the number of months.

62. Example 4 Continued
Total saved
amount
saved
for each
month.
start
amount is
plus
Jared y =
$25 + $5 x
David y =
$40 + $5 x
y = 5x + 25
y = 5x + 40
Both equations are in the slope-intercept form.
y = 5x + 25
y = 5x + 40
The lines have the same slope but different y-intercepts.
The graphs of the two equations are parallel lines, so there is no solution. If the patterns continue, the amount in Jared’s account will never be equal to the amount in David’s account.

63. Check It Out! Example 4
Matt has $100 in a checking account and deposits $20 per month. Ben has $80 in a checking account and deposits $30 per month. Will the accounts ever have the same balance? Explain.
Write a system of linear equations. Let y represent the account total and x represent the number of months.
y = 20x + 100
y = 30x + 80
Both equations are in slope-intercept form.
y = 20x + 100
y = 30x + 80
The lines have different slopes..
The accounts will have the same balance. The graphs of the two equations have different slopes so they intersect.

64. Lesson Quiz: Part I
Solve and classify each system. 1. 2. 3.
y = 5x – 1
infinitely many solutions; consistent, dependent
5x – y – 1 = 0
y = 4 + x
no solution; inconsistent
–x + y = 1
y = 3(x + 1)
consistent, independent
y = x – 2

65. Lesson Quiz: Part II
4. If the pattern in the table continues, when will the sales for Hats Off equal sales for Tops?
never