Beginning Algebra, 11th Edition

Step 3 Add the two equations.

Add in columns.

Step 4 Solve. Divide by 3.

Step 5 Find the value of yby substituting 5 for xin either of the original equations.

(1)

Let.

Multiply.

Subtract 11.

Step 6 Check by substituting and into both of the original equations.

CHECK (1) (2)

Substitute. Substitute.

✓ True ✓ True

Since is a solution of bothequations, the solution set is.

NOW TRY

1 5, - 12 51 5, - 126

10 = 10 25 = 25

1 - 12 + 11  2152 5152 = - 1 + 26

y+ 11 = 2 x 5 x=y+ 26

x= 5 y=- 1

y= - 1

y+ 11 = 10

y+ 11 = 2152 x= 5

y+ 11 = 2 x

x= 5

3 x = 15

5 x-y=^26

- 2 x+y=- 11

266 CHAPTER 4 Systems of Linear Equations and Inequalities

NOW TRY
EXERCISE 3
Solve the system.

2 x+ 8 y=- 6

3 x- 5 y= 25

OBJECTIVE 2 Multiply when using the elimination method.Sometimes

we need to multiply each side of one or both equations in a system by some number

before adding will eliminate a variable.

Using the Elimination Method

Solve the system.

(1)

(2)

Adding the two equations gives which does not eliminate

either variable. However, we can multiply each equation by a suitable number so that

the coefficients of one of the two variables are opposites. For example, to eliminate x,

we multiply each side of (equation (1)) by 5 and each side of

(equation (2)) by.

Multiply equation (1) by 5.

Multiply equation (2) by.

Add.

Divide by 11.

Find the value of xby substituting for yin either equation (1) or (2).

(2)

Let.

Multiply.

Add 14.

Divide by 5.

Check that the solution set of the system is 51 3, - 726. NOW TRY

x= 3

5 x= 15

5 x- 14 = 1

5 x+ 21 - 72 = 1 y=- 7

5 x+ 2 y= 1

- 7

y= - 7

11 y=- 77

-^10 x-^4 y=-^2 -^2

10 x+ 15 y=- 75

5 x+ 2 y= 1 - 2

2 x+ 3 y=- 15

7 x+ 5 y=-14,

5 x+ 2 y= 1

2 x+ 3 y=- 15

EXAMPLE 3

The coefficients of x

are opposites.

NOW TRY
EXERCISE 2
Solve the system.

5 x- 3 y=- 27

2 x- 6 =- 3 y

NOW TRY ANSWERS

2. 51 - 3, 4 26 3. 51 5, - 226

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