# Elimination Method - Part 2 - Multiply or Divide

## Solve System of Equations - Elimination Method - Part 2 - Multiplication - Divide- How it Works - Video

### Example 1

Example 1:

When dealing with a system of equations, there are several ways to solve for it. Here we are going to use the elimination method. So we are either going to add or subtract the two equations together so we can eliminate one variable.

We can add the two equations together when the coefficients of the same variable are opposite like 5 and -5.

We can subtract the two equations together when the coefficients of the same variable are the same like 4 and 4.

Neither variable has coefficients that are the same or opposites. So we have to change one or both for that to happen. We need to use multiplication or division or both to make that happen.

In this case, we multiplied the first equation by 2 to make the coefficients in front of y the same, 8 and 8.

Now we can subtract the two equations since we have 8 and 8 in front of y.

10x - 0y = 40

10x = 40

x = 4

Multiply 1st equation by 2 then subtract the two equations

Drop the 0

Divide both sides by 10

Next we can substitute the x = 4 into either equation to solve for y. We chose 3x + 4y = 4 because it is the first one.

3 * (4) + 4y = 4

12 + 4y = 4

4y = -8

y = -2

Substitute 4 for x

Multiply

Subtract 12 to both sides

Divide both sides by 4

So the answer to the system is (4, -2)

What do you think the graph looks like?

### Example 2

Example 2:

When dealing with a system of equations, there are several ways to solve for it. Here we are going to use the elimination method. So we are either going to add or subtract the two equations together so we can eliminate one variable.

We can add the two equations together when the coefficients of the same variable are opposite like 5 and -5.

We can subtract the two equations together when the coefficients of the same variable are the same like 4 and 4.

Neither variable has coefficients that are the same or opposites. So we have to change one or both for that to happen. We need to use multiplication or division or both to make that happen.

In this case, we multiplied the first equation by 5 and the second equation by 2 to make the coefficients in front of x the opposite, 10 and -10.

Now we can add the two equations since we have 10 and -10 in front of x.

0x + 14y = 28

14y = 28

y = 2

Multiply 1st equation by 5 and 2nd equation by 5 then add the two equations

Drop the 0

Divide both sides by 14

Next we can substitute the y = 2 into either equation to solve for x. We chose 2x + 4y = 6 because it is the first one.

2x + 4 * (2) = 6

2x + 8 = 6

2x = -2

x = -1

Substitute 2 for y

Multiply

Subtract 8 to both sides

Divide both sides by 2

So the answer to the system is (-1, 2)

What do you think the graph looks like?

### Example 3

Example 3:

When dealing with a system of equations, there are several ways to solve for it. Here we are going to use the elimination method. So we are either going to add or subtract the two equations together so we can eliminate one variable.

We can add the two equations together when the coefficients of the same variable are opposite like 5 and -5.

We can subtract the two equations together when the coefficients of the same variable are the same like 4 and 4.

Neither variable has coefficients that are the same or opposites. So we have to change one or both for that to happen. We need to use multiplication or division or both to make that happen.

In this case, we divided the second equation by 2 to make the coefficients in front of x the same, 5 and 5.

Now we can subtract the two equations since we have 5 and 5 in front of x.

0x - 2y = 4

-2y = 4

y = -2

Divide 2nd equation by 2 then subtract the two equations

Drop the 0

Divide both sides by -2

Next we can substitute the y = -2 into either equation to solve for x. We chose 5x + 2y = 6 because it is the first one.

5x + 2 * (-2) = 6

5x - 4 = 6

5x = 10

x = 2

Substitute 2 for y

Multiply

Add 4 to both sides

Divide both sides by 5

So the answer to the system is (2, -2)

What do you think the graph looks like?