Review Substitution Method

y = 5x

y = 2x + 3

Are either equation in the system above in the form of y =...?

y = 5x

y = 2x + 3

We will substitute top equation into the bottom, what does the substitution look like?

After substituting we end up with:

5x = 2x + 3

How do we start solving for x?

After substituting we end up with:

5x = 2x + 3

Let's move variables! But how do we do it?

After substituting we end up with:

5x = 2x + 3

When we move 2x, do we flip the sign or keep it the same?

After moving 2x, we end up:

5x - 2x = 3

3x = 3

Solve for x

We know x = 1 and our original system is:

y = 5x

y = 2x + 3

Substitute x = 1 into either equation to solve for y.

Try solving this system on your own now:

y = 2x -2

y =3x

Lets try a new type of problem

y = 3x + 8

8x + 4y = 12

Are either equation in the system above in the form of y =...?

Lets sub the top equation into the bottom one.

y = 3x + 8

8x + 4y = 12

What is the end result?

8x + 4(3x + 8) = 12

Lets deal with the 4(3x + 8)

After we distribute what do we end up with?

We end up with this!

8x + 12x + 32= 12

When we collect like terms what do we end up with?

We end up with this!

20x + 32 = 12

Solve this two step equation.

We end up with x = -1

Sub it into the original equations to find y value

y = 3x + 8

8x + 4y = 12

Try solving this system on your own now:

y = 2x + 1

3x + 4y = 26