x y

x y

4 17

3 2 9

Objective:

• Solve systems of equations by the elimination method using multiplication and addition.

Assignment:

• pp. 460 #4-26

Previously, we learned how to solve systems of equations by using addition or subtraction which eliminated one of the variables.

2 3 12

3 6

x y

x y

This system of equations could be solved by eliminating the y variable through addition.

3 2 15

3 9

x y

x y

This system of equations could be solved by eliminating the x variable through subtraction.

x y

x y

4 17

3 2 9

Since this system of equations can’t be solved by elimination with addition or elimination with subtraction, how can we solve it?

If the top equation was multiplied by 3, then the first term would be 3x. The bottom equation could then be subtracted from the top equation eliminating the variable x.

3 4 3 17

3 12 51

( ) ( )x y

x y

The new system of equations is now:

3 12 51

3 2 9

x y

x y

3 12 51

3 2 9

x y

x y

Subtract the bottom equation from the top equation.(-)

14 42y

Solve for y.14

14

42

14

y

y 3 Solve for x by substituting the value for y into one of the equations.

3 2 9x y 3 2 3 9x ( )

3 6 9x

+ 6 + 6

3 15x

3

3

15

3

x x 5

x 5 y 3x y

x y

4 17

3 2 9Substitute the value of the variables into each equation to determine if the solution is correct.

x y 4 175 4 3 17 ( )

5 12 17 17 17

3 2 9x y 3 5 2 3 9( ) ( )

15 6 9 9 9

This system of equations represents two lines which intersect at the point (5,3).

x y

x y

4 17

3 2 9

Previously, we solved this system of equations by multiplying the top equation by 3 and then used elimination by subtraction.

Could we have used a different factor for the multiplication?

We could have multiplied the bottom equation by 2 to get

6 4 18x y The system of equations would then become

x y

x y

4 17

6 4 18

Elimination by addition would then be used to solve this system of equations. The result should be the same. Try it and see.

1. Arrange the equations with like terms in columns.

2. Multiply one or both equations by an appropriate factor so that the new coefficients of x or y have the same absolute value.

3. Add or subtract the equations and solve for the remaining variable.

4. Substitute the value for that variable into one of the equations and solve for the value of the other variable.

5. Check the solution in each of the original equations.

Solve the following systems of equations by using elimination.

1 . 4 3 12

2 14

x y

x y

2 . 16 8 12

12 6 9

x y

x y

3 .

2 4 3

3 6 8

x y

x y

4 3 12

2 14

x y

x y

Multiply the bottom equation by 4 to get a new system of equations.

4 3 12

4 8 56

x y

x y

Subtract the bottom equation from the top equation.

(-)

11 44y

Solve for y.

11

11

44

11

y

y 4Solve for x by substituting the value for y into one of the equations.

x y 2 14x 2 4 14( )

x 8 14x 6

4 3 12

2 14

x y

x y

x 6 y 4

Check the solution by substituting the values for the variables into each equation.

4 3 12

4 6 3 4 12

24 12 12

12 12

x y

x y

2 14

6 2 4 14

6 8 14

14 14

This system of equations represents two lines which intersect at the point (6,4).

16 8 12

12 6 9

x y

x y

The lowest common multiple of 6 and 8 is 24. Multiply the top equation by 3 and the bottom equation by 4.

3 16 8 3 12

48 24 36

( ) ( )x y

x y

4 12 6 4 9

48 24 36

( ) ( )x y

x y

The new system of equations is:48 24 36

48 24 36

x y

x y

Since these two equations are identical, there is only one line and an infinite number of solutions.

2 4 3

3 6 8

x y

x y

The lowest common multiple of 2 and 3 is 6. Multiply the top equation by 3 and the bottom equation by 2.

3 2 4 3 3

6 12 9

( ) ( ) x y

x y

2 3 6 2 8

6 12 16

( ) ( )x y

x y

Add the new system of equations together.

6 12 9

6 12 16

x y

x y0 0 25x y

Since 0 ≠ 25, there is no solution to this system of equations which represents two parallel lines.