Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 14.3 Solving Systems of...

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.

14.3

Solving Systems of Linear Equations by Addition

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 22


The Elimination Method

Another method that can be used to solve a system of linear equations is called the addition or elimination method.

The addition method is based on the addition property of equality: Adding equal quantities on both sides of an equation results in an equivalent equation.



Solve the following system of equations using the elimination method.

x + y = 7x – y = 9


Example

Continued

Add the equations to eliminate y.

x + y = 7

x – y = 9

2x = 16

Divide both sides by 2.x = 8



Substitute 8 for x into one of the original equations and solve for y. x + y = 7


Example continued

Our computations have produced the point (8, –1).

Replace the x value with 8 in the first equation.

8 + y = 7

Add 8 to both sides of the equation.y = –1

Continued.



Check the point in the original equations.


Example continued

The solution of the system is (8, –1).

8 + (–1) = 7

x – y = 9x + y = 7

8 – (–1) = 9

True. True.



Solve the following system of equations using the elimination method.

6x – 3y = –34x + 5y = –9

Multiply both sides of the first equation by 5 and the second equation by 3.5(6x – 3y) = 5(–3) 30x – 15y = –15 Add the equations

3(4x + 5y) = 3(–9) 12x + 15y = –27 to eliminate y.


Example

Continued

42x = –42 x = –1 Divide both sides by 42.



Substitute –1 for x into one of the original equations and solve for y. 6x – 3y = –3

6(–1) – 3y = –3 Replace the x value in the first equation.

–6 – 3y = –3 Simplify.

–3y = 3 Add 6 to both sides.

y = –1 Divide both sides by –3.

Our computations have produced the point (–1, –1).


Example continued

Continued



Check the point in the original equations.First equation,

6x – 3y = –3

6(–1) – 3(–1) = –3 True

Second equation,

4x + 5y = –9

4(–1) + 5(–1) = –9 True

The solution of the system is (–1, –1).


Example continued



To Solve a System of Two Linear Equations by the Addition MethodStep 1:Rewrite each equation in standard form, Ax + By = C.Step 2:If necessary, multiply one or both equations by a nonzero number so that the coefficients of a chosen variable in the system are opposites.Step 3:Add the equations.Step 4:Find the value of one variable by solving equation from Step 3.Step 5:Find the value of the second variable by substituting the value found in step 4 into either of the original equations.Step 6:Check the proposed solution in the original system.




Solve the system of equations using the addition method.

24

1

2

12

3

4

1

3

2

yx

yx

First multiply both sides of the equations by a number that will clear the fractions out of the equations.


Example

Continued



Multiply both sides of each equation by 12. (Note: you don’t have to multiply each equation by the same number, but in this case it will be convenient to do so.)

First equation,

2

3

4

1

3

2 yx

2

312

4

1

3

212 yx Multiply both sides by 12.

1838 yx Simplify.


Example continued

Continued



Add the two resulting equations.

8x + 3y = – 18

6x – 3y = – 24

14x = – 42

x = –3 Divide both sides by 14.

Second equation,2

4

1

2

1 yx

2124

1

2

112

yx Multiply both sides by 12.

Simplify.2436 yx


Example continued

Continued



Substitute –3 for x into one of the original equations.

8x + 3y = –188(–3) + 3y = –18

–24 + 3y = –18

3y = –18 + 24 = 6

y = 2

We need to check the ordered pair (–3, 2) in both equations of the original system.


Example continued

Continued



Check the point in the original equations.

First equation:

2

3

4

1

3

2 yx

2 1 3( ) ( )3 4 2

3 2

2

3

2

12 True

Second equation:

24

1

2

1 yx

1 1( ) ( ) 22 4

3 2

22

1

2

3 True

The solution is (–3, 2).


Example continued



In a similar fashion to what you found in the last section, use of the addition method to combine two equations might lead you to results like . . .

5 = 5 (which is always true, thus indicating that there are infinitely many solutions, since the two equations represent the same line), or

0 = 6 (which is never true, thus indicating that there are no solutions, since the two equations represent parallel lines).

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Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 14.3 Solving Systems of...

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