Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems 3-2 Using Algebraic Methods to Solve Linear Systems Holt Algebra 2 Warm Up

Holt McDougal Algebra 2

3-2Using Algebraic Methods to Solve Linear Systems3-2 Using Algebraic Methods

to Solve Linear Systems

Holt Algebra 2

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz



3-2Using Algebraic Methods to Solve Linear Systems

Warm UpDetermine if the given ordered pair is an element of the solution set of

2x – y = 5

3y + x = 6

1. (3, 1) yes 2. (–1, 1) no

Solve each equation for y.

3. x + 3y = 2x + 4y – 4

4. 6x + 5 + y = 3y + 2x – 1

y = –x + 4

y = 2x + 3

.



Solve systems of equations by substitution. Solve systems of equations by elimination.

Objectives



substitutionelimination

Vocabulary



The graph shows a system of linear equations. As you can see, without the use of technology, determining the solution from the graph is not easy. You can use the substitution method to find an exact solution. In substitution, you solve one equation for one variable and then substitute this expression into the other equation.



Use substitution to solve the system of equations.

Example 1A: Solving Linear Systems by Substitution

y = x – 1

x + y = 7Step 1 Solve one equation for one variable. The first equation is already solved for y: y = x – 1.

Step 2 Substitute the expression into the other equation.

x + y = 7

x + (x – 1) = 72x – 1 = 7

2x = 8x = 4

Substitute (x –1) for y in the other equation.

Combine like terms.



Step 3 Substitute the x-value into one of the original equations to solve for y.

Example 1A Continued

y = x – 1

y = (4) – 1

y = 3

Substitute x = 4.

The solution is the ordered pair (4, 3).




Check A graph or table supports your answer.




Example 1B: Solving Linear Systems by Substitution

2y + x = 4

3x – 4y = 7Method 1 Isolate y.

2y + x = 4

5x = 15

3x + 2x – 8 = 7

First equation.

Method 2 Isolate x.

Isolate one variable.

Second equation.

Substitute the expression intothe second equation.

Combine like terms.

2y + x = 4

x = 3

x = 4 – 2y

3x – 4y= 7

3(4 – 2y)– 4y = 7

12 – 6y – 4y = 7

12 – 10y = 7–10y = –55x – 8 = 7

First part of the solution

3x –4 +2 = 7

y = + 2

3x – 4y = 7



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

2y + (3) = 4

2y = 1

Example 1B Continued

Substitute the value to solve for the other variable.

Second part of the solution

2 + x = 4

1 + x = 4

x = 3

By either method, the solution is .

Method 1 Method 2




y = 2x – 1

3x + 2y = 26Step 1 Solve one equation for one variable. The first equation is already solved for y: y = 2x – 1.

Step 2 Substitute the expression into the other equation. 3x + 2y = 26

3x + 2(2x–1) = 26

3x + 4x – 2 = 26

7x = 28x = 4

Substitute (2x –1) for y in the other equation.Combine like terms.

Check It Out! Example 1a




y = 2x – 1

y = 2(4) – 1

y = 7

Substitute x = 4.

The solution is the ordered pair (4, 7).

Check It Out! Example 1a Continued



Check A graph or table supports your answer.




Use substitution to solve the system of equations.5x + 6y = –9

2x – 2 = –y

Method 1 Isolate y.

2x – 2 = –y First equation.

Isolate one variable.

Substitute the expression into the second equation.

5x + 6y = –9

Check It Out! Example 1b

Method 2 Isolate x.

y = –2x + 2

Second equation.

2x – 2 = –y

x = 1 – y

5(1 – y)+ 6y = –9

5x + 6y = –9

5x + 6(–2x + 2) = –9



Combine like terms.

Method 1 Isolate y. Method 2 Isolate x.

–7x = –21

5x + 6(–2x + 2) = –9

x = 3

5x – 12x + 12 = –9

First part of the solution.

Check It Out! Example 1b Continued

10 – 5y + 12y = –18

10 + 7y = –18

7y = –28

y = –4



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

5(3) + 6y = –9

15 + 6y = –9

Example 1b Continued

Substitute the value to solve for the other variable.


5x + (–24) = –9

x = 3

By either method, the solution is (3, –4).

6y = –24

y = –4

5x + 6(–4) = –9

5x = 15



You can also solve systems of equations with the elimination method. With elimination, you get rid of one of the variables by adding or subtracting equations. You may have to multiply one or both equations by a number to create variable terms that can be eliminated.

The elimination method is sometimes called the addition method or linear combination.

Reading Math



Use elimination to solve the system of equations.

Example 2A: Solving Linear Systems by Elimination

3x + 2y = 4

4x – 2y = –18

Step 1 Find the value of one variable.

3x + 2y = 4+ 4x – 2y = –18

The y-terms have opposite coefficients.


7x = –14

x = –2

Add the equations to eliminate y.





3(–2) + 2y = 4

2y = 10

y = 5 Second part of the solution

The solution to the system is (–2, 5).




Example 2B: Solving Linear Systems by Elimination

3x + 5y = –16

2x + 3y = –9

Step 1 To eliminate x, multiply both sides of the first equation by 2 and both sides of the second equation by –3.

Add the equations.

First part of the solutiony = –5

2(3x + 5y) = 2(–16)

–3(2x + 3y) = –3(–9)

6x + 10y = –32

–6x – 9y = 27



Example 2B Continued


3x + 5(–5) = –16

3x = 93x – 25 = –16

x = 3

Step 2 Substitute the y-value into one of the original equations to solve for x.

The solution for the system is (3, –5).



Example 2B: Solving Linear Systems by Elimination

Check Substitute 3 for x and –5 for y in each equation.

3x + 5y = –16 2x + 3y = –9

–163(3) + 5(–5)

–16 –16

2(3) + 3(–5) –9

–9 –9



Use elimination to solve the system of equations. 4x + 7y = –25

–12x –7y = 19


Step 1 Find the value of one variable.

The y-terms have opposite coefficients.


–8x = –6

4x + 7y = –25

– 12x – 7y = 19

Add the equations to eliminate y.

x =




3 + 7y = –25

7y = –28



4( ) + 7y = –25

y = –4

The solution to the system is ( , –4).




5x – 3y = 42

8x + 5y = 28

Step 1 To eliminate x, multiply both sides of the first equation by –8 and both sides of the second equation by 5.

Add the equations.


y = –4


49y = –196

–8(5x – 3y) = –8(42)

5(8x + 5y) = 5(28)

–40x + 24y = –336

40x + 25y = 140




5x – 3(–4) = 42

5x = 305x + 12 = 42

x = 6

Step 2 Substitute the y-value into one of the original equations to solve for x.

The solution for the system is (6,–4).




Check Substitute 6 for x and –4 for y in each equation.

5x – 3y = 42 8x + 5y = 28

425(6) – 3(–4)

42 42

8(6) + 5(–4) 28

28 28




In Lesson 3–1, you learned that systems may have infinitely many or no solutions. When you try to solve these systems algebraically, the result will be an identity or a contradiction.

An identity, such as 0 = 0, is always true and indicates infinitely many solutions. A contradiction, such as 1 = 3, is never true and indicates no solution.

Remember!



Classify the system and determine the number of solutions.

Example 3: Solving Systems with Infinitely Many or No Solutions

3x + y = 1

2y + 6x = –18

Because isolating y is straightforward, use substitution.

Substitute (1–3x) for y in the second equation.

Solve the first equation for y.

3x + y = 1

2(1 – 3x) + 6x = –18

y = 1 –3x

2 – 6x + 6x = –182 = –18

Distribute.

Simplify.

Because 2 is never equal to –18, the equation is a contradiction. Therefore, the system is inconsistent and has no solution.

x



Classify the system and determine the number of solutions. 56x + 8y = –32

7x + y = –4


Substitute (–4 –7x) for y in the first equation.

Solve the second equation for y.

7x + y = –4

56x + 8(–4 – 7x) = –32

y = –4 – 7x

56x – 32 – 56x = –32 Distribute.

Simplify.

Because –32 is equal to –32, the equation is an identity. The system is consistent, dependent and has infinite number of solutions.


–32 = –32



Classify the system and determine the number of solutions. 6x + 3y = –12

2x + y = –6


Substitute (–6 – 2x) for y in the first equation.

Solve the second equation.

2x + y = –6

6x + 3(–6 – 2x)= –12

y = –6 – 2x

6x –18 – 6x = –12 Distribute.

Simplify.

Because –18 is never equal to –12, the equation is a contradiction. Therefore, the system is inconsistent and has no solutions.


–18 = –12 x



A veterinarian needs 60 pounds of dog food that is 15% protein. He will combine a beef mix that is 18% protein with a bacon mix that is 9% protein. How many pounds of each does he need to make the 15% protein mixture?

Example 4: Zoology Application

Let x present the amount of beef mix in the mixture.

Let y present the amount of bacon mix in the mixture.



Example 4 Continued

Write one equation based on the amount of dog food:

Amount of beef mix

plus amount of bacon mix

equals

x y

60.

60+ =

Write another equation based on the amount of protein:

Protein of beef mix

plus protein of bacon mix

equals

0.18x 0.09y

protein in mixture.

0.15(60)+ =



Solve the system.x + y = 60

0.18x +0.09y = 9

x + y = 60

y = 60 – x

First equation

0.18x + 0.09(60 – x) = 9

0.18x + 5.4 – 0.09x = 9

0.09x = 3.6

x = 40


Substitute (60 – x) for y.

Distribute.

Simplify.

Example 4 Continued



Substitute the value of x into one equation.

Substitute x into one of the original equations to solve for y.

40 + y = 60

y = 20 Solve for y.

The mixture will contain 40 lb of the beef mix and 20 lb of the bacon mix.

Example 4 Continued



A coffee blend contains Sumatra beans which cost $5/lb, and Kona beans, which cost $13/lb. If the blend costs $10/lb, how much of each type of coffee is in 50 lb of the blend?

Let x represent the amount of the Sumatra beans in the blend.

Check It Out! Example 4

Let y represent the amount of the Kona beans in the blend.



Write one equation based on the amount of each bean:

Amount of Sumatra beans

plus amount of Kona beans

equals

x y

50.

50+ =

Write another equation based on cost of the beans:

Cost of Sumatra beans plus

cost of Kona beans

equals

5x 13y

cost of beans.

10(50)+ =

Check It Out! Example 4 Continued



Solve the system.x + y = 50

5x + 13y = 500

x + y = 50

y = 50 – x

First equation

5x + 13(50 – x) = 500

5x + 650 – 13x = 500–8x = –150

x = 18.75


Substitute (50 – x) for y.Distribute.

Simplify.




Substitute the value of x into one equation.

Substitute x into one of the original equations to solve for y.

18.75 + y = 50

y = 31.25 Solve for y.

The mixture will contain 18.75 lb of the Sumatra beans and 31.25 lb of the Kona beans.




Lesson Quiz

Use substitution or elimination to solve each system of equations.

3x + y = 1

y = x + 91.

(–2, 7)

5x – 4y = 10

3x – 4y = –22.

(6, 5)

3. The Miller and Benson families went to a theme park. The Millers bought 6 adult and 15 children tickets for $423. The Bensons bought 5 adult and 9 children tickets for $293. Find the cost of each ticket.adult: $28; children’s: $17

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Holt McDougal Algebra 2 3-2 Using Algebraic Methods to Solve Linear Systems 3-2 Using Algebraic Methods to Solve Linear Systems Holt Algebra 2 Warm Up