holt mcdougal algebra 1 5-3 solving systems by elimination 5-3 solving systems by elimination holt...

Holt McDougal Algebra 1

5-3 Solving Systems by Elimination5-3 Solving Systems by Elimination

Holt Algebra 1

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5-3 Solving Systems by Elimination

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.

7.

6.

8.

3 and 6

6 and 8

4 and 10

2 and 5

6

24

20

10



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

Objectives



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.



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.



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 3Substitute the value of the variable into one of the original equations and solve for the other variable.

Step 4Write the answers from Steps 2 and 3 as an ordered pair, (x, y), and check.



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.



Example 1: Elimination Using Addition

3x – 4y = 10x + 4y = –2

Solve by elimination.

Step 1 3x – 4y = 10 Align like terms. −4y and +4y are opposites.

Add the equations to eliminate y.

4x = 8 Simplify and solve for x.

x + 4y = –24x + 0 = 8Step 2

Divide both sides by 4.4x = 84 4x = 2



Example 1 Continued

Step 3 x + 4y = –2 Write one of the original equations.

2 + 4y = –2 Substitute 2 for x.–2 –2

4y = –4

4y –44 4y = –1

Step 4 (2, –1)

Subtract 2 from both sides.

Divide both sides by 4.

Write the solution as an ordered pair.



Check It Out! Example 1

y + 3x = –22y – 3x = 14


Align like terms. 3x and −3x are opposites. Step 1

2y – 3x = 14y + 3x = –2

Add the equations to eliminate x.Step 2 3y + 0 = 12

3y = 12 Simplify and solve for y.

Divide both sides by 3.

y = 4



Step 3 y + 3x = –2

Check It Out! Example 1 Continued

Write one of the original equations.

4 + 3x = –2 Substitute 4 for y.Subtract 4 from both sides.–4 –4

3x = –6Divide both sides by 3.3x = –6

3 3x = –2

Write the solution as an ordered pair.Step 4 (–2, 4)



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.



2x + y = –52x – 5y = 13


Example 2: Elimination Using Subtraction

Both equations contain 2x. Add the opposite of each term in the second equation.

Step 1–(2x – 5y = 13)

2x + y = –5

2x + y = –5–2x + 5y = –13

Eliminate x.

Simplify and solve for y.

0 + 6y = –18 Step 26y = –18

y = –3



Example 2 Continued


Step 3 2x + y = –5

2x + (–3) = –5Substitute –3 for y.

Add 3 to both sides.2x – 3 = –5

+3 +3

2x = –2 Simplify and solve for x.

x = –1


Step 4 (–1, –3)



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

Remember!




3x + 3y = 15–2x + 3y = –5


3x + 3y = 15 –(–2x + 3y = –5)

Step 1

3x + 3y = 15+ 2x – 3y = +5

Both equations contain 3y. Add the opposite of each term in the second equation.

Eliminate y.

Simplify and solve for x.

5x + 0 = 20

5x = 20

x = 4

Step 2





Substitute 4 for x.


Step 3 3x + 3y = 15

3(4) + 3y = 15

12 + 3y = 15–12 –12

3y = 3

y = 1Simplify and solve for y.


(4, 1)Step 4



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.



x + 2y = 11–3x + y = –5

Solve the system by elimination.

Example 3A: Elimination Using Multiplication First

Multiply each term in the second equation by –2 to get opposite y-coefficients.

x + 2y = 11Step 1 –2(–3x + y = –5)

x + 2y = 11+(6x –2y = +10) Add the new equation to

the first equation to eliminate y.7x + 0 = 21

Step 2 7x = 21

x = 3Solve for x.



Example 3A Continued


Step 3 x + 2y = 11

Substitute 3 for x. 3 + 2y = 11Subtract 3 from both sides.–3 –3

2y = 8y = 4

Solve for y.


Step 4 (3, 4)



–5x + 2y = 32

2x + 3y = 10


Example 3B: Elimination Using Multiplication First

Step 1 2(–5x + 2y = 32) 5(2x + 3y = 10)

Multiply the first equation by 2 and the second equation by 5 to get opposite x-coefficients –10x + 4y = 64

+(10x + 15y = 50) Add the new equations to eliminate x.

Solve for y.

19y = 114

y = 6

Step 2



Example 3B Continued


Step 3 2x + 3y = 10

Substitute 6 for y. 2x + 3(6) = 10

Subtract 18 from both sides.–18 –18

2x = –8

2x + 18 = 10

x = –4 Solve for x.

Step 4 Write the solution as an ordered pair.

(–4, 6)



Check It Out! Example 3a


3x + 2y = 6

–x + y = –2

Step 1 3x + 2y = 6 3(–x + y = –2)

3x + 2y = 6 +(–3x + 3y = –6)

0 + 5y = 0

Multiply each term in the second equation by 3 to get opposite x-coefficients.

Add the new equation to the first equation.

Simplify and solve for y.5y = 0

y = 0

Step 2



Check It Out! Example 3a Continued


Step 3 –x + y = –2

Substitute 0 for y. –x + 3(0) = –2

–x + 0 = –2

–x = –2

Solve for x.


(2, 0)

x = 2



Check It Out! Example 3b


2x + 5y = 26

–3x – 4y = –25

Step 1 3(2x + 5y = 26) +(2)(–3x – 4y = –25)

Multiply the first equation by 3 and the second equation by 2 to get opposite x-coefficients 6x + 15y = 78

+(–6x – 8y = –50) Add the new equations to eliminate x.

Solve for y. y = 40 + 7y = 28Step 2



Check It Out! Example 3b Continued


Step 3 2x + 5y = 26

Substitute 4 for y. 2x + 5(4) = 26

Solve for x.


(3, 4) x = 3

2x + 20 = 26–20 –20

2X = 6




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 + 0.75c = 7.75 The cost of felt and card stock totals $7.75.

f + c = 12 The total number of sheets is 12.



Example 4 Continued

Step 1 0.50f + 0.75c = 7.75

+ (–0.50)(f + c) = 12

Multiply the second equation by –0.50 to get opposite f-coefficients.

0.50f + 0.75c = 7.75+ (–0.50f – 0.50c = –6)

Add this equation to the first equation to eliminate f.

Solve for c.

Step 2 0.25c = 1.75

c = 7

Step 3 f + c = 12

Substitute 7 for c.f + 7 = 12

–7 –7f = 5






Step 4 (7, 5)

Paige can buy 7 sheets of card stock and 5 sheets of felt.

Example 4 Continued




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 + 0.90t = 14.85 The cost of lilies and tulips totals $14.85.

l + t = 13 The total number of flowers is 13.




Step 1 1.25l + .90t = 14.85+ (–.90)(l + t) = 13

Multiply the second equation by –0.90 to get opposite t-coefficients.

1.25l + 0.90t = 14.85 + (–0.90l – 0.90t = –11.70) Add this equation to the

first equation to eliminate t.

Solve for l.Step 2

0.35l = 3.15

l = 9





Step 4 (9, 4)

Sally bought 9 lilies and 4 tulips.

Step 3 Write one of the original equations.

Substitute 9 for l.9 + t = 13

–9 –9t = 4


l + t = 13



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



Lesson QuizSolve each system by elimination.

1.

2.

3.

(2, –3)

(11, 3)

(–3, –7)

2x + y = 253y = 2x – 13

x = –2y – 4–3x + 4y = –18

–2x + 3y = –153x + 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


5-3 Solving Systems by Elimination5-4 Solving Special Systems

Holt Algebra 1

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz




Warm UpSolve 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

4. y = 3x + 22x + y = 7

Solve by using any method.

(1, 5) 5. x – y = 8x + y = 4

(6, –2)



Solve special systems of linear equations in two variables.

Classify systems of linear equations and determine the number of solutions.

Objectives



inconsistent systemconsistent systemindependent systemdependent system

Vocabulary



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.



Example 1: Systems with No Solution

Method 1 Compare slopes and y-intercepts.

y = x – 4 y = 1x – 4 Write both equations in slope-intercept form.

–x + y = 3 y = 1x + 3

Show that has no solution.y = x – 4

–x + y = 3

The lines are parallel because they have the same slope and different y-intercepts.

This system has no solution.



Example 1 Continued

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.



–x + y = 3



Example 1 Continued

Check Graph the system.

The lines appear are parallel.

– x + y = 3

y = x – 4


–x + y = 3





Show that has no solution.y = –2x + 5

2x + y = 1

y = –2x + 5 y = –2x + 5

2x + y = 1 y = –2x + 1

Write both equations in slope-intercept form.

The lines are parallel because they have the same slope and different y-intercepts.




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.

False.


5 = 1



2x + y = 1



Check Graph the system.

The lines are parallel.

y = – 2x + 1

y = –2x + 5



2x + y = 1



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.



Show that has infinitely many solutions.

y = 3x + 2

3x – y + 2= 0

Example 2A: Systems with Infinitely Many Solutions


y = 3x + 2 y = 3x + 2 Write both equations in slope-intercept form. The lines have the same slope and the same y-intercept.

3x – y + 2= 0 y = 3x + 2

If this system were graphed, the graphs would be the same line. There are infinitely many solutions.



Method 2 Solve the system algebraically. Use the elimination method.

y = 3x + 2 y − 3x = 2

3x − y + 2= 0 −y + 3x = −2

Write equations to line up like terms.

Add the equations.

True. The equation is an identity.

0 = 0

There are infinitely many solutions.

Example 2A Continued


y = 3x + 2

3x – y + 2= 0



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

Caution!





y = x – 3

x – y – 3 = 0


y = x – 3 y = 1x – 3 Write both equations in slope-intercept form. The lines have the same slope and the same y-intercept.

x – y – 3 = 0 y = 1x – 3

If this system were graphed, the graphs would be the same line. There are infinitely many solutions.



Method 2 Solve the system algebraically. Use the elimination method.

Write equations to line up like terms.

Add the equations.

True. The equation is an identity.

y = x – 3 y = x – 3

x – y – 3 = 0 –y = –x + 3

0 = 0

There are infinitely many solutions.



y = x – 3

x – y – 3 = 0



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.



Example 3A: Classifying Systems of Linear Equations

Solve3y = x + 3

x + y = 1

Classify the system. Give the number of solutions.


3y = x + 3 y = x + 1

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.



Example 3B: Classifying Systems of Linear equations

Solvex + y = 5

4 + y = –x


x + y = 5 y = –1x + 5

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.



Example 3C: Classifying Systems of Linear equations


Solvey = 4(x + 1)

y – 3 = x

y = 4(x + 1) y = 4x + 4

y – 3 = x y = 1x + 3


The lines have different slopes. They intersect.

The system is consistent and independent. It has one solution.



Check It Out! Example 3a


Solvex + 2y = –4

–2(y + 2) = x


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.



Check It Out! Example 3b


Solvey = –2(x – 1)

y = –x + 3

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.



Check It Out! Example 3c


Solve2x – 3y = 6

y = x

y = x y = x

2x – 3y = 6 y = x – 2 Write both equations in slope-intercept form.

The lines have the same slope and different y-intercepts. They are parallel.

The system is inconsistent. It has no solutions.



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.



Total saved is

startamount plus

amountsaved

for eachmonth.

Jared y = $25 + $5 x

David y = $40 + $5 x

Both equations are in the slope-intercept form.

The lines have the same slope but different y-intercepts.

y = 5x + 25y = 5x + 40

y = 5x + 25y = 5x + 40

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.

Example 4 Continued



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 + 100y = 30x + 80

y = 20x + 100y = 30x + 80

Both equations are in slope-intercept form.

The lines have different slopes..

The accounts will have the same balance. The graphs of the two equations have different slopes so they intersect.



Lesson Quiz: Part I

Solve and classify each system.

1.

2.

3.

infinitely many solutions; consistent, dependent

no solution; inconsistent

y = 5x – 15x – y – 1 = 0

y = 4 + x

–x + y = 1

y = 3(x + 1)y = x – 2

consistent, independent



Lesson Quiz: Part II

4. If the pattern in the table continues, when will the sales for Hats Off equal sales for Tops?

never

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