Question

Find the values of $a\&b$ such that it satisfies following linear equations:
$\frac{4}{a+b}+\frac{2}{a-b}=5\frac{1}{a+b}+\frac{1}{a-b}=3$

• A. $a=\frac{-11}{2}\&b=\frac{7}{20}$
• B. $b=\frac{11}{2}\&a=\frac{-7}{20}$
• C. $a=\frac{-8}{7}\&b=\frac{-6}{7}$
• D. $b=\frac{6}{5}\&a=\frac{-8}{5}$

Answer 1

The correct option is C $a=\frac{-8}{7}\&b=\frac{-6}{7}$

Given, $\frac{4}{a+b}+\frac{2}{a-b}=5\frac{1}{a+b}+\frac{1}{a-b}=3$

Lets take $x=\frac{1}{a+b};y=\frac{1}{a-b}$

$4x+2y=5----\left(i\right)x+y=3----\left(ii\right)$

Multiply $\left(ii\right)$ by $2$
$2x+2y=6----\left(iii\right)$

Substract $\left(iii\right)$ from $\left(i\right)$
$4x+2y=5\underset{–}{{}_{-}2x\pm 2y{=}_{-}6}2x=-1\Rightarrow x=\frac{-1}{2}$

Substitute $x$ in $\left(ii\right)$
$\frac{-1}{2}+y=3\Rightarrow y=\frac{7}{2}\therefore \frac{1}{a+b}=\frac{-1}{2};\frac{1}{a-b}=\frac{7}{2}2=-1(a+b)\Rightarrow -a-b=2---\left(iv\right)2=7(a-b)\Rightarrow 7a-7b=2---\left(v\right)$

Multiply $\left(iv\right)$ with $7$
We get $-7a-7b=14---\left(vi\right)$

Add $\left(v\right)\&\left(vi\right)$
$7a-7b=2\underset{–}{-7a-7b=14}-14b=16\Rightarrow b=\frac{-8}{7}$

Substitute $b$ in $\left(v\right)$
$7a-7\times \frac{-8}{7}=2\Rightarrow a=\frac{-6}{7}$