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\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}

Solve the following similtaneous eQuation,

\sf{\:\dfrac{1}{x}+\dfrac{1}{y}\:=\:8.....(1)} \\ \\ \sf{\:\dfrac{4}{x}-\dfrac{2}{y}\:=\:2......(2)}

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

Let,

\mapsto\sf{\green{\:\dfrac{1}{x}\:=\:p}} \\ \\ \mapsto\sf{\green{\:and\:\dfrac{1}{y}\:=\:q}}

So, new equation will be

  • p + q = 8 .........(3)
  • 4p - 2q = 2 ......(4)

Multiply by 4 in equ(3),

  • 4p + 4q = 32 .......(5)

Subtract equ(5)-equ(4),

\mapsto\sf{\:(4q+2q)\:=\:32-2} \\ \\ \mapsto\sf{\:6q\:=\:30} \\ \\ \mapsto\sf{\:q\:=\:\cancel{\dfrac{30}{6}}} \\ \\ \mapsto\sf{\red{\:q\:=\:5}}

Now, keep value of q in equ(3),

\mapsto\sf{\:(p+5)\:=\:8} \\ \\ \mapsto\sf{\:p\:=\:8-5} \\ \\ \mapsto\sf{\red{\:p\:=\:3}}

\Large{\underline{\mathfrak{\bf{\pink{Thus}}}}}

\mapsto\sf{\:value\:of\:x\:will\:be\:=\:\dfrac{1}{x}\:=\:p} \\ \\ \mapsto\sf{\:value\:of\:x\:will\:be\:=\:\dfrac{1}{x}\:=\:3} \\ \\ \mapsto\sf{\pink{\:value\:of\:x\:will\:be\:=\:\dfrac{1}{3}}}

And,

\mapsto\sf{\:value\:of\:y\:will\:be\:=\:\dfrac{1}{y}\:=\:q} \\ \\ \mapsto\sf{\:value\:of\:y\:will\:be\:=\:\dfrac{1}{y}\:=\:5} \\ \\ \mapsto\sf{\pink{\:value\:of\:y\:will\:be\:=\:\dfrac{1}{5}}}

\Large{\underline{\mathfrak{\bf{\pink{Verification}}}}}

Keep value of 1/x and 1/y in equ(1),

\mapsto\sf{\:\dfrac{1}{x}+\dfrac{1}{y}\:=\:8} \\ \\ \mapsto\sf{\:\dfrac{1}{\dfrac{1}{3}}+\dfrac{1}{\dfrac{1}{5}}\:=\:8} \\ \\ \mapsto\sf{\:3+5\:=\:8} \\ \\ \mapsto\sf{\:8\:=\:8}

L.H.S = R.H.S

THAT'S PROVED


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