Question
Solve the following simultaneous equations.
$\frac{2}{x}+\frac{2}{3 y}=\frac{1}{6} ; \frac{3}{x}+\frac{2}{y}=0$
$\frac{2}{x}+\frac{2}{3 y}=\frac{1}{6} ; \frac{3}{x}+\frac{2}{y}=0$
✓
Answer
Let $\frac{1}{x}=m$ and $\frac{1}{y}=n$
$2 m+\frac{2}{3} n=\frac{1}{6} \Rightarrow 12 m+\frac{12}{3 n}=1 \Rightarrow 12 m+4 n=1 \ldots(1)$
$3 m+2 n=0 \ldots(2)$
Multiply Eq. II by 2
$6 n+4 n=0 \ldots(3)$
Subtract Eq.III from Eq. I
$\begin{aligned}
& 12 m+4 n=1 \\
& -6 m-4 n=0 \\
& \hline 6 m=1 \\
& m=\frac{1}{6}
\end{aligned}$
Substitute $m=1 / 6$ in Eq. I
$\begin{aligned}
& 12 \times \frac{1}{6}+4 n=1 \\
& 2+4 n=1 \\
& 4 n=1-2
\end{aligned}$
$\begin{aligned}
& 4 n =-1 \\
& n =-\frac{1}{4} \\
& \therefore m =\frac{1}{ x } \Rightarrow \frac{1}{6}=\frac{1}{ x } \Rightarrow x =6 \\
& \therefore n =\frac{1}{ y } \Rightarrow-\frac{1}{4}=\frac{1}{ y } \Rightarrow y =-4
\end{aligned}$
Hence, $(x, y)=(6,-4)$
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