# How do you find the solution set for 2y=5x+10 and y^2-4y=5x+10?

##### 1 Answer

#### Answer:

*or*

#### Explanation:

Your system of equations looks like this

Right from the start, you can say that this system of equations can be written as *one equation* in

#(5x+10) = 2y" "# and#" "5x+10 = y^2 - 4y#

This is equivalent to

#underbrace(y^2 - 4y)_(color(blue)(=5x+10)) = overbrace(2y)^(color(orange)(=5x+10))#

Rearrange this equation into classic quadratic form

#y^2 - 4y - 2y = 0#

#y^2 -6y = 0#

You can factor this equation to get

#y * (y - 6) = 0#

The two solutions will thus be

Use these values of

*when*#y=o# ,*you have*

#5x+10 = 0^2 - 4 * 0#

#5x+10 = 0#

#5x = -10 implies x= (-10)/5 = -2#

*when*#y=6# ,*you have*

#5x+10 = 6^2 - 4 * 6#

#5x + 10 = 12#

#5x = 2 implies x = 2/5#

The two solution sets for this system of equations are

*or*