1. SOLVING SYSTEMS USING ELIMINATION
September 28/29, 2011

2. Elimination Method eliminate
The goal is to _____________ one of the variables by _________ your equations.
adding

3. Elimination Method - Steps
Step 1: MULTIPLY one or both of the equations by a constant to obtain coefficients that differ only in sign for one of the variables.
Step 2: ADD the revised equations from Step 1. Combining like terms will eliminate one of the variables. Solve for the remaining variables.
Step 3: SUBSTITUTE the value obtained in Step 2 into either of the original equations and solve for the other variable.

4. Elimination Method –Example 1
x - 2y = x + 2y = 45
First, look to see if any terms can cancel out. In this case, they can! Which one, x or y? y!
4x - 2y = x + 2y = 45 +
5x + 0 = 125
5x = 125
x = 25
Now that you know x = 25, plug it back in to either equation to solve for y:
4x - 2y = 80
-2y = -20
y = 10
4(25) – 2y = 80
100 – 2y = 80
The solution is (25, 10)

5. Elimination Method –Example 2
x – 2y = 20
3x + 3y = 18
In this case, neither variable can be cancelled out. Which term is going to be easier to work with? x!
6x – 2y = 20
3x + 3y = 18
-2( )
To cancel out the x, we will need to multiply the second equation by a “-2”.
You should get….
6x – 2y = 20
-6x - 6y = -36
Now, you can add them like on the previous example!

6. Elimination Method –Example 2
6x – 2y = 20
-6x - 6y = -36 +
y = -16
-8y = -16
y = 2
Now that you know y = 2 plug it back in to either equation to solve for x:
3x + 3y = 18
3x + 3 (2) = 18
3x + 6 = 18
3x = 12
x = 4
The solution to the systems of equations is (4, 2)

7. Elimination Method –Example 3
x + 6y = 28
3x + 4y = 27
By looking at the variables, neither the x or the y is going to be easy to cancel. In this case, we will have to multiply both equations by something! Let’s cancel out the x…
2 and 3 are both factors of … 6
2x + 6y = 28
3x + 4y = 27
So let’s multiply the first equation by 3 and the bottom by -2.
3 ( )
-2 ( )

8. Elimination Method –Example 3
2x + 6y = 28
3x + 4y = 27
3 ( )
-2 ( )
6x + 18y = 84
-6x – 8y = -54 +
y = 30
10y = 30
y = 3
Now that you know y = 3 plug it back in to either equation to solve for x:
2x + 6y = 28
2x + 6 (3) = 28
The solution is (5, 3)
x = 5

9. Elimination Method –Word Problem
In one week, a music store sold 9 guitars for a total of $ Electric guitars sold for $479 each and acoustic guitars sold for $339 each. How many of each type of guitar were sold?
Step 1: Identify your variables
x = number of electric guitars
y = number of acoustic guitars
Step 2: Write a systems of equations
x + y = 9
479x + 339y = 3611

10. Elimination Method –Word Problem
In one week, a music store sold 9 guitars for a total of $ Electric guitars sold for $479 each and acoustic guitars sold for $339 each. How many of each type of guitar were sold?
Step 3: Solve the systems using elimination.
x y =
479x + 339y = 3611
-479( )
-479x – 479y = -4311
479x + 339y = 3611 +
The solution is 4 electric and 5 acoustical guitars!
0x - 140y = -700
y = 5….and x = 4