the student will be able to: solve systems of equations using elimination with multiplication....

The student will be able to:

solve systems of equations using elimination with

multiplication.

Objective

So far, we have solved systems using graphing, substitution, and elimination.

This goes one step further and show how to use ELIMINATION with multiplication.

What happens when the coefficients are not the same?

We multiply the equations to make them the same! You’ll see…

Solving Systems of Equations

1) Solve the system using elimination.

2(2) + 2y = 6

4 + 2y = 6

2y = 2

y = 1

2x + 2y = 6

3x – y = 5Multiply the bottom

equation by 2

2x + 2y = 6

(2)(3x – y = 5)

2x + 2y = 6(+) 6x – 2y = 10 8x = 16 x = 2

(2, 1)

x + 4y = 7

4x – 3y = 9

2) Solve the system using elimination.

Multiply the top equation by -4

(-4)(x + 4y = 7)

4x – 3y = 9

-4x – 16y = -28 (+) 4x – 3y = 9

-19y = -19 y = 14x – 3y = 9

4x – 3(1) = 9

4x – 3 = 9

4x = 12

x = 3

(3, 1)

Which variable is easier to eliminate?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32

3x + y = 44x + 4y = 6

y is easier to eliminate. Multiply the 1st equation by (-4)

3x + 4y = -14x – 3y = 7

3) Solve the system using elimination.

3(1) + 4y = -1

3 + 4y = -1

4y = -4

y = -1

Multiply both equations

(3)(3x + 4y = -1)

(4)(4x – 3y = 7)

9x + 12y = -3 (+) 16x – 12y = 28

25x = 25 x = 1

(1, -1)

4) What is the best number to multiply the top equation by to eliminate the x’s?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32

3x + y = 46x + 4y = 6

Multiply the 1st equation by (-2) (-2)(3x + y = 4) -6x -2y = -8

6x + 4y = 6

2y = -2

y = 1

6x + 4y = 66x + 4(1) = 6 6x = 2

x = 2/6x = 1/3

(1/3, 1)

5) Solve using elimination.

2x – 3y = 1x + 2y = -3

Multiply 2nd equation by (-2) to eliminate the x’s. (-2)(x + 2y = -3) -2x – 4y = 6

1st equation 2x – 3y = 1

-7y = 7

y = -1

2x – 3y = 1

2x – 3(-1) = 1

2x + 3 = 1

2x = -2

x = -1(-1, -1)

2(4x + y = 6 ) -8x – 2y = 13

8x + 2y = 12 -8x – 2y = 13

0 = 25

Add down to eliminate x.

But look what happens, y is eliminated too.

We now have a false statement, thus the system has no solution, it is inconsistent.

5) Solve using elimination.

6) Elimination

Solve the system using elimination.

5x – 2y = –15 3x + 8y = 37

20x – 8y = –60 3x + 8y = 37

23x = –23 x = –1

Since neither variable will drop out if the equations are added together. Multiply one or both ofthe equations by a constant to make one of the variables have the same number with opposite signs.

The best choice is to multiply the top equation by 4 since only one equation would have to be multiplied. Also, the signs on the y-terms arealready opposites.

(4)

6) Elimination

3x + 8y = 37 (second equation)

3(–1) + 8y = 37–3 + 8y = 37 8y = 40 y = 5

The solution is (–1, 5)

To find the second variable, substitute in any equation that contains two variables.

x = –1

7) Elimination

Solve the system using elimination.

4x + 3y = 8 3x – 5y = –23

20x + 15y = 40 9x – 15y= –69

29x = –29 x = –1

For this system, we must multiply both equations by a different constant in order to make one of the variables “drop out.”

It would work to multiply the top equation by –3 and the bottom equation by 4 OR to multiply the top equation by 5 and the bottom equation by 3.

(5)(3)

7) Elimination

4x + 3y = 84(–1) + 3y = 8

–4 + 3y = 8 3y = 12 y = 4

The solution is (–1, 4)

x = –1

8) Elimination Method

Solve the system.

To eliminate x, multiply equation (1) by –2 and

equation (2) by 3 and add the resulting equations.

3696

286

yx

yx

2

3417

y

y

1232

143

yx

yx (1)

(2)

8) Elimination Method

Substitute 2 for y in (1) or (2).

The solution is (3, 2)3

93

1)2(43

x

x

x

9) Solving an Inconsistent System

Solve the system

Eliminate x by multiplying (1) by 2 and adding the result to (2).

Solution set is .

746

423

yx

yx (1)

(2)

746

846

yx

yx

150 Inconsistent System

10) Solving a System with Dependent Equations

Solve the system.

Eliminate x by multiplying (1) by 2 and adding the result to (2).

Each equation is a solution of the other.

Infinite solutions.

42824

yxyx (1)

(2)

428

428

yx

yx

00