Question

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Answer 1

⛦Equation Given :-

• $\leadsto\sf\dfrac{3+x^2}{8+x^2}\sf{=}\dfrac{-3}{4}$

⛦To find :-

• The positive value of the variable.

⛦Solution :-

• Let us consider x² be = y

A/Q

$\dashrightarrow\sf\dfrac{3+y}{8+y}\sf{=}\dfrac{-3}{4}$

$\dashrightarrow\sf{4(3-y)=-3(8+y)}$

${\blue{\text{\sf{[By cross Multiplication]}}}}$

$\dashrightarrow\sf{12-4y=-24+(-3y)}$

$\dashrightarrow\sf{12-4y=-24-3y}$

$\dashrightarrow\sf{-4y+3y=-24-12}$

$\dashrightarrow\sf{\cancel{-}y=\cancel{-}36}$

$\dashrightarrow{\underline{\boxed{\sf{\pink{y=36}}}}}$

$\sf{Thus,y=x^2=36}$

$\mapsto{\sf{x^2=36}}$

$\mapsto\sf{x=\sqrt{36}}$

$\mapsto\sf{x=\sqrt{6\times6}}$

$\mapsto\sf{x=6}$

Therefore, the positive value of the variable is :$\sf\red{6}$

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Answer 2

Answer:

Answer

$\dashrightarrow\sf\dfrac{3+y}{8+y}\sf{=}\dfrac{-3}{4}$

$\dashrightarrow\sf{4(3-y)=-3(8+y)}$

$\dashrightarrow\sf{12-4y=-24+(-3y)}$

$\dashrightarrow\sf{12-4y=-24-3y}$

$\dashrightarrow\sf{-4y+3y=-24-12}$

$\dashrightarrow\sf{\cancel{-}y=\cancel{-}36}$

$\dashrightarrow{\underline{\boxed{\sf{\orange{y=36}}}}}$

$\sf{y=x^2=36}$

$\mapsto{\sf{x^2=36}}$

$\mapsto\sf{x=\sqrt{36}}$

$\mapsto\sf{x=\sqrt{6\times6}}$

$\mapsto\sf{x=6}$

Therefore, the positive value of the variable 6