3.2 Solving Linear Systems Algebraically

• What are the steps to solve a system by substitution?• What clue will you see to know if substitution is a

good choice?• What are the steps to solve a system by linear

combination?• How many solutions are possible for a linear

system?

USING ALGEBRAIC METHODS TO SOLVE SYSTEMS

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

THE SUBSTITUTION METHOD

1

2

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

Solve one of the equations for one of its variables.

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

Solve the linear system using the substitution method.

3 x + 4y – 4 Equation 1 x + 2y 2 Equation 2

x + 2y 2 x – 2y + 2

The Substitution Method

3x + 4y – 43(– 2y + 2) + 4y – 4

y 5

Write Equation 2.

Revised Equation 2.

Substitute – 2y + 2 for x.

Write Equation 1.

Solve Equation 2 for x.

Simplify.

Substitute the expression for x into Equation 1 and solve for y.

SOLUTION

The Substitution Method

x – 2y + 2

x – 2(5) + 2

x – 8

Write revised Equation 2.

Substitute 5 for y.

Simplify.

The solution is (– 8, 5).

Substitute the value of y into revised Equation 2 and solve for x.

Solve the linear system using the substitution method.

3 x + 4y – 4 Equation 1 x + 2y 2 Equation 2

3x + 4y – 43(– 2y + 2) + 4y – 4

y 5Substitute – 2y + 2 for x.

Write Equation 1.

Simplify.

Substitute the expression for x into Equation 1 and solve for y.

The Substitution Method

Check the solution by substituting back into the original equation.

3x + 4y – 4 x + 2y 2

– 4 – 4 Solution checks. 2 2

3 (– 8) + 4 (5) – 4?

– 8 + 2 (5) 2?

Write original equations.

Substitute x and y.

Solve the linear system using the substitution method.

3 x + 4y – 4 Equation 1 x + 2y 2 Equation 2

CHECK

USING ALGEBRAIC METHODS TO SOLVE SYSTEMS

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.

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.

THE LINEAR COMBINATION METHOD

USING ALGEBRAIC METHODS TO SOLVE SYSTEMS

1

2

3 Substitute value obtained in Step 2 into either original equation and solve for other variable.

Multiply one or both equations by a constant to obtain coefficients that d iffer only in sign for one of the variables.

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

Solve the linear system using thelinear combination method.

2 x – 4y 13 Equation 14 x – 5y 8 Equation 2

– 4x + 8y – 26

4 x – 5y 8

The Linear Combination Method: Multiplying One Equation

2 x – 4y 13

4 x – 5y 8

Add the revised equations and solve for y.

3y –18

y – 6

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

SOLUTION

• – 2

The Linear Combination Method: Multiplying One Equation

2 x – 4y 13

2 x – 4(– 6) 13

2 x + 24 13

x – 112

The solution is – , – 6 .( 112 )

y – 6Add the revised equations and solve for y.

Write Equation 1.

Substitute – 6 for y.

Simplify.

Solve for x.

Substitute the value of y into one of the original equations.

Solve the linear system using thelinear combination method.

2 x – 4y 13 Equation 14 x – 5y 8 Equation 2

You can check the solution algebraically using the method shown in the previous example.

CHECK

The Linear Combination Method: Multiplying Both Equations

7 x – 12 y – 22 Equation 1– 5 x + 8 y 14 Equation 2

Solve the linear system using thelinear combination method.

7 x – 12 y – 22

– 5 x + 8 y 14

14 x – 24y – 44

– 15 x + 24y 42

Add the revised equations and solve for x.

– x – 2

x 2

Multiply the first equation by 2 and the second equation by 3 so that the coefficients of y differ only in sign.

SOLUTION

• 2

• 3

The Linear Combination Method: Multiplying Both Equations

– 5 x + 8 y 14

y = 3

– 5 (2) + 8 y 14

The solution is (2, 3).

x 2Add the revised equations and solve for x.

Write Equation 2.

Substitute 2 for x.

Solve for y.

Substitute the value of x into one of the original equations. Solve for y.

7 x – 12 y – 22 Equation 1– 5 x + 8 y 14 Equation 2

Solve the linear system using thelinear combination method.

Check the solution algebraically or graphically.

Linear Systems with Many or No Solutions

x – 2 y 32 x – 4 y 7

Solve the linear system

x – 2 y 3

x 2 y + 3

Solve the first equation for x.

Since the coefficient of x in the first equation is 1, use substitution.

SOLUTION

Linear Systems with Many or No Solutions

x – 2 y 32 x – 4 y 7

Solve the linear system

2 x – 4 y 7

2(2 y + 3) – 4 y 7

6 7

Write second equation.

Substitute 2 y + 3 for x.

Simplify.

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

Substitute the expression for x into the second equation.

x 2 y + 3 Solve the first equation for x.

Linear Systems with Many or No Solutions

6 x – 10 y 12 – 15 x + 25 y – 30

Solve the linear system

6 x – 10 y 12

– 15 x + 25 y – 30

30 x – 50 y 60

– 30 x + 50 y – 60

0 0Add the revised equations.

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

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

SOLUTION

• 5

• 2

Questions to answer:• What are the steps to solve a system by

substitution?• What clue will you see to know if substitution is

a good choice?• What are the steps to solve a system by linear

combination?• How many solutions are possible for a linear

system?

Assignment 3.2Page 152, 11-47 odd