Solve the following pair of equations by reducing them to a pair of linear equations: 1(3x+y)+1(3x-y)=34; 12(3x+y)-12(3x-y)=-18.

# Solve the following pair of equations by reducing them to a pair of linear equations: .

1. A
$x=2,y=7$
2. B
$x=5,y=0$
3. C
$x=1,y=1$
4. D
$x=2,y=3$

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### Solution:

Given equations are, .
Let $\frac{1}{\left(3x+y\right)}=a$ and $\frac{1}{\left(3x-y\right)}=b$.
Then, $a+b=\frac{3}{4}$
$⇒4a+4b=3$………….(1)
Also, $\frac{a}{2}-\frac{b}{2}=\frac{-1}{8}$.
$\frac{a-b}{2}=\frac{-1}{8}$
$a-b=\frac{-1}{4}$
$4a-4b=-1$……………..(2)
Adding equation (1) and equation (2) we get,
$8a=2$
$a=\frac{1}{4}$
Substituting $a=\frac{1}{4}$ in equation (1) we get,

$4b=2$
$b=\frac{1}{2}$
Then, $\frac{1}{3x+y}=\frac{1}{4}$.
$3x+y=4$…………..(3)
And $\frac{1}{3x-y}=\frac{1}{2}$.
$3x-y=2$………………(4)
Adding equation (3) and equation (4)  we get,
$6x=6$
$x=1$
Substituting $x=1$ in equation (3) we get,
$3×1+y=4$
$y=1$
Therefore, option (3) is correct.

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