Question

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

Answer 1

`{(x = -2), (y=0):}" "` or `" "{(x=2/5), (y=6):}`

Explanation:

Your system of equations looks like this

`{(2y = 5x + 10), (y^2 - 4y = 5x+10):}`

Right from the start, you can say that this system of equations can be written as one equation in `y`, since you have

`(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 `y=0` and `y=6`.

Use these values of `y` in one of the two original equations to get the values of `x`.

• 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

`{(x = -2), (y=0):}" "` or `" "{(x=2/5), (y=6):}`