Question

please anssssssssssssss​

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{\pink{Given}}}}}$

$\mapsto\sf{\green{\:\dfrac{3}{x+y}+\dfrac{2}{x-y}\:=\:2...(1)}} \\ \\ \mapsto\sf{\green{\:\dfrac{9}{x+y}-\dfrac{4}{x-y}\:=\:1....(2)}}$

$\underline{\sf{\pink{\:Step-by-Step-Explanation}}}$

First,Let

$\mapsto\sf{\:\dfrac{1}{x+y}\:=\:p} \\ \\ \mapsto\sf{\:\dfrac{1}{x-y}\:=\:q}$

New, equation,

• 3p + 2q = 2 .....(3)
• 9p - 4q = 1 .......(4)

Multiply by 9 in (3) and 3 in (4) ,

• 27p + 18q = 18
• 27p - 12q = 3

_________________subtract it

$\mapsto\sf{\:30q\:=\:15} \\ \\ \mapsto\sf{\:q\:=\:\dfrac{\cancel{15}}{\cancel{30}}} \\ \\ \mapsto\sf{\pink{\:q\:=\:\dfrac{1}{2}}}$

Keep value of q in equ(3),

$\mapsto\sf{\:3p+2\times \dfrac{1}{2}\:=\:2} \\ \\ \mapsto\sf{\:3p\:=\:2-1} \\ \\ \mapsto\sf{\pink{\:p\:=\:\dfrac{1}{3}}}$

Now, find value of x and y ,

First, for x :-

$\mapsto\boxed{\sf{\:p\:=\:\dfrac{1}{x+y}}} \\ \\ \mapsto\sf{\:\dfrac{1}{x+y}\:=\:\dfrac{1}{3}} \\ \\ \mapsto\sf{\orange{\:x+y\:=\:3....(5)}}$

Again, For y:-

$\mapsto\boxed{\sf{\:q\:=\:\dfrac{1}{x-y}}} \\ \\ \mapsto\sf{\:\dfrac{1}{2}\:=\:\dfrac{1}{x-y}} \\ \\ \mapsto\sf{\orange{\:x-y\:=\:2....(6)}}$

Addition of equ(5) and (6),

$\mapsto\sf{\:2x\:=\:5} \\ \\ \mapsto\sf{\green{\:x\:=\:\dfrac{5}{2}}}$

Keep value of x in equ(5)

$\mapsto\sf{\:\dfrac{5}{2}+y\:=\:3} \\ \\ \mapsto\sf{\:y\:=\:3\:-\:\dfrac{5}{2}} \\ \\ \mapsto\sf{\:y\:=\:\dfrac{6-5}{2}} \\ \\ \mapsto\sf{\green{\:y\:=\:\dfrac{1}{2}}}$

$\Large{\underline{\mathfrak{\bf{\red{\:Thus}}}}}$

$\mapsto\sf{\orange{\:Value\:of\:x\:=\:\dfrac{5}{2}}} \\ \\ \mapsto\sf{\orange{\:Value\:of\:y\:=\:\dfrac{1}{2}}}$

$\underline{\bf{\pink{\:Answer\:Verification}}}$

keep value of x and y in equ(5)

$\mapsto\sf{\:\dfrac{5}{2}+\dfrac{1}{2}\:=\:3} \\ \\ \mapsto\sf{\:\dfrac{5+1}{2}\:=\:3} \\ \\ \mapsto\sf{\:\dfrac{\cancel{6}}{\cancel{2}}\:=\:3} \\ \\ \mapsto\sf{\green{\:3\:=\:3}} \\ \\ \mathfrak{\bf{\:\:\:\:L.H.S.\:=\:R.H.S.}}$

That's proved