Question
Solve the following simultaneous equation.
$\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} $
$\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{1}{8}$
$\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} $
$\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{1}{8}$
✓
Answer
$\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} $
$\frac{1}{[2(3 x+y)]}-\frac{1}{[2(3 x-y)]}=\frac{1}{8} $
$\text { Let } \frac{1}{3 x+y}=m \text { and } \frac{1}{3 x-y}=n$
$m + n =\frac{3}{4} \Rightarrow 4(m+ n )=3 \Rightarrow 4 m+4 n =3 \ldots \text { (I) }$
$\frac{1}{2} m-\frac{1}{2} n =\frac{1}{8} \Rightarrow 8(m- n )=1 \times 2 \Rightarrow 8 m-8 n =2 \ldots \text { (II) }$
Multiply Eq. I by 2
$8 m+8 n =6 \ldots( a )$
$8 m-8 n =2 \ldots \ldots$ (b)Add (a) and (b) to get,
$8 m+8 n+8 m-8 n=8 $
$\Rightarrow 16 m=8$
$m=\frac{8}{16} $
$m=\frac{1}{2} $
$8 \times \frac{1}{2}-8 n=2 $
$\Rightarrow 4-8 n=2 \Rightarrow-8 n=2-4$
$\Rightarrow-8 n=-2 $
$\Rightarrow 8 n =2 $
$\Rightarrow n =\frac{2}{8} $
$\Rightarrow n =\frac{1}{4}$
$\therefore m =\frac{1}{2(3 x + y )} $
$\Rightarrow \frac{1}{2(3 x+y)}=\frac{1}{2}$
$\Rightarrow 2=2(3 x+y)$
$\Rightarrow 2=6 x+2 y \ldots ( I I I)$
$\therefore n=\frac{1}{2(3 x-y)} $
$\Rightarrow \frac{1}{2(3 x-y)}=\frac{1}{4} $
$\Rightarrow 4=2(3 x-y) $
$\Rightarrow 4=6 x-2 y \ldots(I V)$
Add Eq. III and IV
$6 x+2 y=2 $
$\underline{6 x-2 y=4}$
${12 x=6} $
$x=\frac{6}{12} $
$x=\frac{1}{2}$
Substituting $x =\frac{1}{2}$ in Eq. III
$6 \times \frac{1}{2}+2 y=2$
$\Rightarrow 3+2 y=2 $
$\Rightarrow 2 y=2-3 $
$\Rightarrow 2 y=-1 $
$y=-\frac{1}{2}$
Hence $(x, y)=\left(\frac{1}{2},-\frac{1}{2}\right)$
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