# Solve the system: 9x^2 + 4(y-2)^2 = 36 x^2 = 2y

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To solve the system

`9x^2+4(y-2)^2=36`

`x^2=2y`

we need to substitute the second equation into the first, then solve the subsequent quadratic equation.

Upon substitution, we see that

`9(2y)+4(y-2)^2=36` now expand

`18y+4(y^2-4y+4)=36` expand again

`18y+4y^2-16y+16=36` collect like terms

`4y^2+2y-20=0` divide by 2

`2y^2+y-10=0` factor left side

`(2y+5)(y-2)=0` now solve to get:

`y=2` or `y=-5/2` .

Now we need to substitute the y-values into the original equations, but since `x^2=2y` , this means that `y>0` , so the only possible solution is `y=2` .

Now solving for x, we get

`x^2=2(2)=4`

so

`x=+-2.`

**This means there are two solutions `x=2` , `y=2` and `x=-2` , `y=2` . As a check, note that `9(+-2)+4(2-2)^2=36` .**