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nalin [4]
3 years ago
9

Please help solve this system of equations

Mathematics
1 answer:
stepan [7]3 years ago
8 0

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

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