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 $\frac{1}{ y }= n _{2 m}+\frac{2}{3} n =\frac{1}{6} \Rightarrow 12 m+\frac{12}{3 n }=1 \Rightarrow 12 m+4 n =1 \ldots (I)$
$3 m+2 n =0 \ldots (II)$
Multiply Eq. II by 2
$6 n+4 n=0 \ldots (III)$
Subtract Eq.III from Eq. I
$12 m+4 n=1$
$\underline{-6 m-4 n=0} $
$6 m=1$
$m=\frac{1}{6}$
Substitute $m =1 / 6$ in Eq. I
$12 \times \frac{1}{6}+4 n =1 $
$2+4 n =1$
$4 n =1-2 $
$4 n =-1 $
$n =-\frac{1}{4} $
$\therefore m =\frac{1}{ x } \Rightarrow \frac{1}{6}=\frac{1}{ x } \Rightarrow x =6 $
$\therefore n =\frac{1}{ y } \Rightarrow-\frac{1}{4}=\frac{1}{ y } \Rightarrow y =-4$
Hence, $(x, y)=(6,-4)$
$3 m+2 n =0 \ldots (II)$
Multiply Eq. II by 2
$6 n+4 n=0 \ldots (III)$
Subtract Eq.III from Eq. I
$12 m+4 n=1$
$\underline{-6 m-4 n=0} $
$6 m=1$
$m=\frac{1}{6}$
Substitute $m =1 / 6$ in Eq. I
$12 \times \frac{1}{6}+4 n =1 $
$2+4 n =1$
$4 n =1-2 $
$4 n =-1 $
$n =-\frac{1}{4} $
$\therefore m =\frac{1}{ x } \Rightarrow \frac{1}{6}=\frac{1}{ x } \Rightarrow x =6 $
$\therefore n =\frac{1}{ y } \Rightarrow-\frac{1}{4}=\frac{1}{ y } \Rightarrow y =-4$
Hence, $(x, y)=(6,-4)$
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