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

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Answer 1 · 2015-08-11T11:46:35

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

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 $x$ .

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

$5x+10 = 0$

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

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

$5x + 10 = 12$

$5x = 2 implies x = 2/5$

The two solution sets for this system of equations are

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