Question
Solve :$\frac{3}{2 x}+\frac{2}{3 y}=-\frac{1}{3}, \frac{3}{4 x}+\frac{1}{2 y}=-\frac{1}{8}$
✓
Answer
Given equations are $\frac{3}{2 x}+\frac{2}{3 y}=-\frac{1}{3}$ and $\frac{3}{4 x}+\frac{1}{2 y}=-\frac{1}{8}$
Let $\frac{1}{x}=u$ and $\frac{1}{y}=v$
Then, the system of equations become
$\frac{3}{2} u+\frac{2}{3} v=-\frac{1}{3}$ and $\frac{3}{4} u+\frac{1}{2} v=-\frac{1}{8}$
$ \Rightarrow \frac{9 u+4 v}{6}=-\frac{1}{3}$ and $\frac{3 u+2 v}{4}=-\frac{1}{8}$
$\Rightarrow 27 u+12 v=-6$ and $24 u+16 v=-4$
$ \Rightarrow 27 u+12 v+6=0$ and $24 u+16 v+4=0$
$\Rightarrow \frac{u}{12 \times 4-16 \times 6}=\frac{-v}{27 \times 4-24 \times 6}=\frac{1}{27 \times 16-24 \times 12}$
$ \Rightarrow \frac{u}{48-96}=\frac{-v}{108-144}=\frac{1}{432-288}$
$ \Rightarrow \frac{u}{-48}=\frac{-v}{-36}=\frac{1}{144}$
$ \Rightarrow \frac{u}{-48}=\frac{v}{36}=\frac{1}{144}$
$ \Rightarrow u=\frac{-48}{144}=\frac{1}{3}$ and $v=\frac{36}{144}=\frac{1}{4}$
$ \Rightarrow \frac{1}{x}=-\frac{1}{3}$ and $\frac{1}{y}=\frac{1}{4}$
$ \Rightarrow \mathrm{x}=-3$ and $\mathrm{y}=4 .$
Let $\frac{1}{x}=u$ and $\frac{1}{y}=v$
Then, the system of equations become
$\frac{3}{2} u+\frac{2}{3} v=-\frac{1}{3}$ and $\frac{3}{4} u+\frac{1}{2} v=-\frac{1}{8}$
$ \Rightarrow \frac{9 u+4 v}{6}=-\frac{1}{3}$ and $\frac{3 u+2 v}{4}=-\frac{1}{8}$
$\Rightarrow 27 u+12 v=-6$ and $24 u+16 v=-4$
$ \Rightarrow 27 u+12 v+6=0$ and $24 u+16 v+4=0$
$\Rightarrow \frac{u}{12 \times 4-16 \times 6}=\frac{-v}{27 \times 4-24 \times 6}=\frac{1}{27 \times 16-24 \times 12}$
$ \Rightarrow \frac{u}{48-96}=\frac{-v}{108-144}=\frac{1}{432-288}$
$ \Rightarrow \frac{u}{-48}=\frac{-v}{-36}=\frac{1}{144}$
$ \Rightarrow \frac{u}{-48}=\frac{v}{36}=\frac{1}{144}$
$ \Rightarrow u=\frac{-48}{144}=\frac{1}{3}$ and $v=\frac{36}{144}=\frac{1}{4}$
$ \Rightarrow \frac{1}{x}=-\frac{1}{3}$ and $\frac{1}{y}=\frac{1}{4}$
$ \Rightarrow \mathrm{x}=-3$ and $\mathrm{y}=4 .$
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