Question

SYSTEMS OF EQUATIONS

Answer 1

Solve for x and y:
\[8xe^y-8x^3=0\]\[4x^2e^y-e^4=0\]

Answer 2

This is a nonlinear system of two equations and two unknowns.

Answer 3

By the way, I love your avatar.

Answer 4

Okay, I was able to solve it myself.

Let (1) be equation 1, and (2) be equation 2.

Factor (1) as such:\[x(e^y-x^2)=0\] Now x can't be 0, or else (2) won't hold. So e^y must be equal to x^2.

Plug this into (2) to get:\[4x^4-e^4=0\]\[x^4=\frac{e^4}{4}\]\[x=\pm\frac{e}{\sqrt{2}}\]

CASE A: x = +e/sqrt(2)
\[\frac{e}{\sqrt{2}}e^y-\frac{e^3}{2\sqrt{2}}=0\]\[e^y=\frac{e^2}{2}\]\[\ln e^y=\ln \frac{e^2}{2}\]\[y = \ln (e^2) - \ln (2)\]\[y = 2 \ln(e)- \ln(2)\]\[y = 2 - \ln(2)\]

Answer 5

CASE B: x = -e/sqrt(2)

Basically the same as CASE A.