# Solving Systems of Equations by Elimination

In previous Lessons, we learned how to solve systems of equations by graphing our lines and looking for the intersect and then went on to use our properties of equality to manipulate the equations within a system to solve for the solution.

## Example 1

This example is a reminder of why the math we will do in this lesson is correct.

3 + 4 = 7 this is a true equation

1 + 9 = 10 this is a true equation

Now following our properties of equality, we can add these two equations together.

3 + 4 = 7

+(1 + 9) = 10

3 + (1) + 4 + (9) = 7 + (10)

17 = 17, this is also a true equation! We followed the rules. We can use these rules to solve equations.

The solutions in the above animation are repeated with a more detailed explanation below.

## Question 1

Solve the system using elimination and do so twice - eliminating each variable in turn.

8x + 10y =12

-2x + y = 4

Solve by first eliminating variable x.

As before, when solving these systems, we look to multiply one or both of the equations to eliminate either our x or y terms when the equation is added so that we can then solve for the x or y term that is remaining. For these questions we are being told to eliminate x.

Remember our Properties of Equality!

So, we look to multiply both or one of the equations so that the x-terms are cancelled out when we add them together.

Here we see 8x in the first equation and -2x in the second. For these terms to eliminate themselves when added, we can multiply our second equation with the -2x by 4, this will give us 8x and -8x which will eliminate each other when we add them together.

So, multiplying the bottom equation by 4,

8x + 10y =12

+ 4 (-2x + y = 4)

8x + 10y =12

+ -8x + 4y = 16

Then adding the two equations together

8x + (-8x) + 10y + 4y = 12 + 16

As we see, 8x and -8x eliminate each other

0 + 14y = 28

14y = 28

Divide both sides by 14

14y ÷ 14 = 28 ÷ 14

y = 2,

Now we have our y-term we need to find our x-term.

Sub y = 2 into -2x + y = 4

-2x + 2 = 4

Subtract 2 from both sides

-2x + 2 - 2 = 4 - 2

-2x = 2

Divide both sides by -2

-2x ÷ -2 = 2 ÷ -2

x = -1

So, a solution to our set is x = -1 and y = 2, or (-1, 2)

Next, solve by eliminating y first

8x + 10y =12

-2x + y = 4

Here we must eliminate our y-term first, so look at the 10y in the first equation and the y in the second. What would we have to multiply one or both of our equations by, so that when the two equations are added, the two y-terms cancel out?

If we multiply the second equation by -10, then when we would be left with 10y in the first equation and -10y in the second. When the two equations are added together, 10y and -10y would eliminate each other.

So, Multiply the whole of the bottom equation by -10

8x + 10y =12

+(-10) (-2x + y = 4)

Be careful multiplying out the negative!

8x + 10y =12

+(20x + -10y = -40)

8x + 20x + 10y - 10y = 12 - 40

The y-terms cancel out as planned

28x - 0 = -28

Divide both sides by 28

28x ÷ 28 = -28 ÷ 28

x = -1

Just as in a)!

Now sub x = -1 back into either of the equation s to get our y value.

Sub x = - 1 into -2x + y = 4

-2(-1) + y = 4

2 + y = 4

Subtract 2 from both sides

2 + y - 2 = 4 - 2

y = 2

Again, we have the solution, x = -1 and y = 2, (-1, 2)

Note we have the same solution from two different paths. They are both the same because we used correct mathematics and followed the Properties of Equality. The methods applied may have been different, but the methods were both true, meaning the equations remained true and left us with the same answers.

The solutions in the above animation are repeated with a more detailed explanation below.

## Question 2

We have two points (-3, 11) and (1, 7) which a straight line passes through. Recall the general equation for a straight-line y = mx + b. Substitute our x and y values from the points given into this equation, creating two equations. Use elimination to solve for m and b, then state the equation for our line that passes through those points.

This question may seem obscure at first, but with a couple methods we are familiar with, we can create our two equations to make a system of equations which we can solve by elimination.

We want to sub each of our points into the general y = mx + b equation, giving us two equations.

Sub in (-3, 11) or x = -3 y = 11 into y = mx + b

(11) = m (-3) + b,

-3m + b = 11

Now Sub in (1, 7) or x = 1 and y = 7 into y = mx + b

(7) = m (1) + b

m + b = 7

Notice we know have a system of equations, this time with the terms m and b instead of x and y.

-3m + b = 11

m + b = 7

We can now eliminate one of our terms, as we have b in one equation and b in the other, let's choose these terms to eliminate first.

With the Properties of Equality, we can add equations together, but we can also subtract them, in this case, subtracting the top equation by the bottom equation can save us a step of multiplying the bottom equation by - 1 in order for the b-terms to cancel out.

However, if you are more comfortable with the process of adding the equations together, you can multiply the bottom or top equations by -1 and then add, but here we will subtract the top and bottom.

Subtracting the top and bottom equations;

-3m + b = 11

-(m + b = 7)

-3m - m + b - b = 11 - 7

-4m = 4

Divide both sides by -4

-4m ÷ -4 = 4 ÷ -4

m = -1

Now sub our m = - 1 into one of the equations we created, sub m = -1 into m + b = 7

-1 + b = 7