Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide 1.


Systems of Linear Equations and Inequalities

Chapter 4


4.3

Solving Systems of Linear Equations by Elimination


Objectives

1. Solve linear systems by elimination.

2. Multiply when using the elimination method.

3. Use an alternative method to find the second value in a solution.

4. Use the elimination method to solve special systems.

4.3 Solving Systems of Linear Equations by Elimination


This addition can be taken a step further. Adding equal quantities, rather than the same quantity, to both sides of an equation also results in equal sums.

Solving Linear Systems by Elimination

If A = B and C = D, then A + C = B + D.


Using the addition property to solve systems is called the elimination method. When using this method, the idea is to eliminate one of the variables. To do this, one of the variables in the two equations must have coefficients that are opposites.


Example 1

Use the elimination method to solve the system.

2x – y = 5 x + y = 7

Each equation in this system is a statement of equality, so the sum of the left sides equals the sum of the right sides.



+

3x = 12Notice that y has been eliminated. Now we can solve the result for x.

3 3x = 4

To find the y-value of the solution, substitute 4 for x into either equation.

x + y = 7

4 + y = 7–4 –4

y = 3 The solution set of the system is {(4, 3)}.




CAUTIONA system is not completely solved until values for both x and y are found. Do not stop after finding the value of only one variable. Remember to write the solution set as a set containing an ordered pair.


Solving a Linear System by Elimination

Step 1 Write both equations in standard form Ax + By = C.Step 2 Transform so that the coefficients of one pair of

variable terms are opposites. Multiply one or both equations by appropriate numbers so that the sum of the coefficients of either the x- or y-terms is 0.

Step 3 Add the new equations to eliminate a variable. The sum should be an equation with just one variable.

Step 4 Solve the equation from Step 3 for the remaining variable.Step 5 Substitute the result from Step 4 into either of the original

equations and solve for the other variable.Step 6 Check the solution in both of the original equations. Then

write the solution set.




Example 2

Solve the system.

Step 1 Both equations are already in standard form.

Multiplying When Using the Elimination Method


3x + y = 8 5x – 2y = 6

Step 2 The y-terms have opposite signs, but their coefficients are different. However, if we multiply both sides of the first equation by 2, then the y terms will cancel when we add.

2( )

6x + 2y = 16 5x – 2y = 6

Step 3 Now when we add, the y terms are eliminated.

+

11x = 22Step 4 We are left with an equation with just x, which we can solve easily.

11 11

x = 2

( )2


Example 2 (concluded)

Solve the system.

Step 5 Substitute x = 2 into either of the equations.

Multiplying When Using the Elimination Method


3x + y = 8 5x – 2y = 6

The solution set is {(2, 2)}.

3(2) + y = 8

6 + y = 8–6 –6

y = 2

Step 6 Check that (2,2) satisfies both equations.

3(2) + 2 = 8 ?

6 + 2 = 8 ?

8 = 8

5(2) + – 2(2) = 6 ?

10 – 4 = 6 ?

6 = 6


x + 4y = 9 3x + 2y = 5

–3( )

Example 3

Solve the system.

We multiply the first equation by –3 to eliminate the x’s.

Using an Alternative Method to Find a Second Value


x + 4y = 9 3x = –2y + 5

( )·–3

–3x – 12y = –27 3x + 2y = 5

+

–10y = –2210 10

11

5y


Solve the system.


Substituting –11/5 for y into one of the given equations would give x, but the arithmetic involved would be messy. Instead, solve for x by starting again with the original equations and eliminating y.

Using an Alternative Method to Find a Second Value


11

5y

x + 4y = 9 3x + 2y = 5

–2( ) ( )·–2

x + 4y = 9 –6x – 4y = –10 +

–5x = –15 5

1

5x 1 11

The solution set is , .5 5


Example 4

Solve each system by the elimination method.

Multiply each side of the first equation by 3; then add the two equations.

Using the Elimination Method to Solve Special Systems


A true statement occurs when the equations are equivalent. As before, this indicates that every solution of one equation is also a solution of the other. The solution set is

(a) ⅔x – y = 1 –2x + 3y = –3

3( ) ( )3 2x – 3y = 3 –2x + 3y = –3 +

0 = 0 True.

3 2 3 , .x y x y


(b) x + 4y = 3 2x + 8y = 5

–2( )


Solve each system by the elimination method.

Using the Elimination Method to Solve Special Systems


( )·–2 –2x – 8y = –6 2x + 8y = 5

+

0 = –1

The false statement 0 = –1 indicates that the solution set is

False.

0 .

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide 1.

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