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

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To solve the system



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 




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` .

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