3 .1 Solve Linear Systems by Graphing

1 / 22

# 3 .1 Solve Linear Systems by Graphing - PowerPoint PPT Presentation

3 .1 Solve Linear Systems by Graphing. 3.1 Solve Linear Systems by Graphing. 3.1 Solve Linear Systems by Graphing. 3.1 Solve Linear Systems by Graphing. U SING A LGEBRAIC M ETHODS TO S OLVE S YSTEMS. 1. 2. 3. THE SUBSTITUTION METHOD.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## 3 .1 Solve Linear Systems by Graphing

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

USING ALGEBRAIC METHODS TO SOLVE SYSTEMS

1

2

3

THE SUBSTITUTION METHOD

In this lesson you will study two algebraic methods for solving linear systems. The first method is called substitution.

Solve one of the equations for one of its variables.

Substitute expression from Step 1 into other equation and solve for other variable.

Substitute value from Step 2 into revised equation from Step 1. Solve.

The Substitution Method

3x + 4y  – 4Equation 1

x + 2y  2Equation 2

Solve the linear system using

the substitution method.

SOLUTION

Solve Equation 2 forx.

x + 2y  2

Write Equation 2.

x –2y + 2

Revised Equation 2.

Substitute the expression forxinto Equation 1 and solve for y.

3x + 4y  – 4

Write Equation 1.

Substitute – 2y + 2for x.

3(–2y + 2) + 4y  – 4

y5

Simplify.

The Substitution Method

Substitute the expression forxinto Equation 1 and solve fory.

3x + 4y  – 4

Write Equation 1.

3(–2y + 2) + 4y  – 4

Substitute –2y + 2for x.

y5

Simplify.

3x + 4y  – 4Equation 1

x + 2y  2Equation 2

Solve the linear system using

the substitution method.

Substitute the value ofyinto revised Equation 2 and solve forx.

x –2y+ 2

Write revised Equation 2.

Substitute 5for y.

x –2(5) + 2

x–8

Simplify.

The solution is (–8, 5).

The Substitution Method

?

?

3 (–8) + 4 (5)  –4

–8 + 2 (5)  2

CHECK

3x + 4y  – 4Equation 1

x + 2y  2Equation 2

Solve the linear system using

the substitution method.

Check the solution by substituting back into the original equation.

3x + 4y  – 4

x + 2y  2

Write original equations.

Substitute x and y.

Solution checks.

–4  –4

2  2

USING ALGEBRAIC METHODS TO SOLVE SYSTEMS

CHOOSING A METHOD In the first step of the previous example, you could have solved for either x or y in either Equation 1 or Equation 2. It was easiest to solve for x in Equation 2 because the x-coefficient was 1. In general you should solve for a variable whose coefficient is 1 or –1.

If neither variable has a coefficient of 1 or –1, you can still use substitution. In such cases, however, the linear combination method may be better. The goal of this method is to add the equations to obtain an equation in one variable.

USING ALGEBRAIC METHODS TO SOLVE SYSTEMS

THE LINEAR COMBINATION METHOD

1

2

3

Multiply one or both equations by a constant to

obtain coefficients that differ only in sign for one

of the variables.

Add revised equations from Step 1. Combine like terms to eliminate one of the variables. Solve for remaining variable.

Substitute value obtained in Step 2 into either

original equation and solve for other variable.

The Linear Combination Method: Multiplying One Equation

2x – 4y  13Equation 1

4x – 5y  8Equation 2

Solve the linear system using thelinear combination method.

SOLUTION

Multiply the first equation by–2so that x-coefficients differ only in sign.

• –2

2x –4y  13

–4x+8y  – 26

4x –5y  8

4x– 5y  8

3y  –18

and solve fory.

y–6

The Linear Combination Method: Multiplying One Equation

Add the revised equations and solve fory.

y–6

You can check the solution algebraically using

the method shown in the previous example.

CHECK

11

x –

2

(

)

11

The solution is –, –6 .

2

2x – 4y  13Equation 1

4x – 5y  8Equation 2

Solve the linear system using thelinear combination method.

Substitute the value ofyinto one of the original equations.

2x – 4y  13

Write Equation 1.

Substitute –6for y.

2x – 4(–6)  13

2x + 24  13

Simplify.

Solve for x.

The Linear Combination Method: Multiplying Both Equations

7x – 12y  –22Equation 1

– 5x + 8y  14Equation 2

Solve the linear system using thelinear combination method.

SOLUTION

Multiplythe first equation by2and the second equation by3so that the coefficients ofydiffer only in sign.

• 2

7x – 12y  –22

14x–24y  –44

–15x+ 24y  42

• 3

–5x + 8y  14

–x–2

Addthe revised equations and solve for x.

x 2

The Linear Combination Method: Multiplying Both Equations

x 2

Add the revised equations and solve for x.

7x – 12y  –22Equation 1

– 5x + 8y  14Equation 2

Solve the linear system using thelinear combination method.

Substitute the value of xinto one of the original equations. Solve for y.

–5x + 8y  14

Write Equation 2.

Substitute 2for x.

–5 (2) + 8y  14

y= 3

Solve for y.

The solution is (2, 3).

Check the solution algebraically or graphically.

Linear Systems with Many or No Solutions

x – 2y  3

2x – 4y  7

Solve the linear system

SOLUTION

Since the coefficient ofxin the first equation is 1, use substitution.

Solve the first equation for x.

x – 2y  3

x 2y+ 3

Linear Systems with Many or No Solutions

Solve the first equation for x.

x 2y+ 3

x – 2y  3

2x – 4y  7

Solve the linear system

Substitute the expression forxinto the second equation.

2x – 4y  7

Write second equation.

2(2y+ 3) – 4y  7

Substitute 2y+ 3forx.

6  7

Simplify.

Because the statement 6 = 7 is never true, there is no solution.

Linear Systems with Many or No Solutions

6x – 10y  12 –15x + 25y  –30

Solve the linear system

SOLUTION

Since no coefficient is 1 or –1, use the linear combination method.

30x– 50y 60

•5

6x – 10y  12

–30x+ 50y –60

•2

–15x + 25y  –30

0  0

Because the equation 0 = 0 is always true, there are infinitely many solutions.

Animated Activity Online

SPECIAL TYPES OF SOLUTIONS:

http://www.classzone.com/cz/books/algebra_2_2007_na/resources/applications/animations/explore_learning/chapter_3/dswmedia/7_5_special_sys.html

### Homework:

Pg. 143-144 #43-45, 56

Pg. 152-153 #12-14, 25-27, 35, 36, 54