gvaldytist

2022-06-29

Explain method to solve two equations in two unknowns where one of the variables has a square term

Given:

$-\frac{96}{{x}^{2}y}+1+y=0$

$-\frac{96}{x{y}^{2}}+2+x=0$

Solve for $x$ and $y$

How should I find $x$ and $y$?

Given:

$-\frac{96}{{x}^{2}y}+1+y=0$

$-\frac{96}{x{y}^{2}}+2+x=0$

Solve for $x$ and $y$

How should I find $x$ and $y$?

Christina Ward

Beginner2022-06-30Added 19 answers

Well, we can start doing this:

$\frac{96}{{x}^{2}y}=1+y$

$\frac{96}{x{y}^{2}}=2+x$

Multiply the top equation by $x$ and the bottom by $y$, equate the right hand sides, and subtract out $xy$ gives $x=2y$. Substituting back into the top equation and rearranging gives

${y}^{4}+{y}^{3}={y}^{3}(y+1)=24,$

which has solution $y=2$. Then, $x=4$.

$\frac{96}{{x}^{2}y}=1+y$

$\frac{96}{x{y}^{2}}=2+x$

Multiply the top equation by $x$ and the bottom by $y$, equate the right hand sides, and subtract out $xy$ gives $x=2y$. Substituting back into the top equation and rearranging gives

${y}^{4}+{y}^{3}={y}^{3}(y+1)=24,$

which has solution $y=2$. Then, $x=4$.

Eden Solomon

Beginner2022-07-01Added 7 answers

I think that's expert's answer is wrong, what do you mean?

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