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SYSTEMS OF EQUATIONS - QuestionCove
OpenStudy (anonymous):

SYSTEMS OF EQUATIONS

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (across):

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

5 years ago
OpenStudy (across):

By the way, I love your avatar.

5 years ago
OpenStudy (anonymous):

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) \[\frac{e}{\sqrt{2}}e^y-\frac{e^3}{2\sqrt{2}}=0\]\[e^y=\frac{e^2}{2}\]\[\ln e^y=\ln \frac{e^2}{2}\]\[y = \ln (e^2) - \ln (2)\]\[y = 2 \ln(e)- \ln(2)\]\[y = 2 - \ln(2)\]

5 years ago
OpenStudy (anonymous):

CASE B: x = -e/sqrt(2) Basically the same as CASE A.

5 years ago
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