If 3x*3y=y*x^2+96 and y=x+8, what is the value of y?

Question

Please give me the answer and the steps to find the answer (srry i'm giving this question im just bored)

Answer 1

Since we're given y = x+8, let's solve the first equation for x by plugging in y.

3x*3y=y*x^2+96 <==> 3x*3(x+8)=(x+8)*x^2+96
Move everything to one side and expand.
(x+8)*x^2+96-3x*3(x+8) = 0
x^3+8x^2 -9x^2 -24x + 96 = 0
x^3 - x^2 -24x + 96 = 0

This is easiest solved by plugging in to the solver of a graphic calculator. Once you get values for x, plug in to y = x + 8 and you have values for y.

Answer 2

3x*3y=y*x^2+96
y=x+8 substitute

3x*3(x+8)=(x+8)x^2+96 distribute where necessary
9x^2+72x=x^3+8x^2+96 subtract LHS from both sides
0=x^3-x^2-72x+96 there is at leasr 1 real root. 1 will be somewhere between 1 & 2. around 1.35
y will be x+8= around 9.35

Answer 3

3x*3y = y*x^2 +96
y=x+8
x = y - 8
9y(y - 8) = y(y - 8)^2 + 96
9y^2 - 72y = y(y^2 - 16y + 64) + 96
9y^2 - 72y = y^3 - 16y^2 + 64y + 96
y^3 - 25y^2 + 136y + 96 = 0
y ≈ -0.630873981

Answer 4

3x*3(x+8)=(x+8)*x^2+96
9x^2+72x=x^3+8x^2+96
x^3-x^2-72x+96=0
solve to get x and substitute in y=x+8 to get y
the values of x are -8.63 or 8.29
so y will be -0.63 or +16.29