Question

Solve this question immediately
No spam allow​

Answer 1

$\large{\underline{\underline{\red{\sf{Given:}}}}}$

• $\tt{\dfrac{2}{\sqrt{x}}+\dfrac{3}{\sqrt{y}}=2}$
• $\tt{\dfrac{4}{\sqrt{x}}+\dfrac{9}{\sqrt{y}}=1}$

$\large{\underline{\underline{\red{\sf{To\:Find:}}}}}$

• $\tt{Value\:of\:x\:and\:y.}$
• $\tt{We\:have\:to\:use\: elimination\: method.}$

$\large{\underline{\underline{\red{\sf{Answer:}}}}}$

$\underline{\purple{\bf{\dag Elimination\: Method:-}}}$

$\tt{Here\:are\:the\:steps.}$

$\tt{\green{Step \:1:-}\:\:\orange{Multiply\:variable\:with\:a\:number\:to\:make\:same\: coefficients.}}$

$\tt{\green{Step\:2:-}\orange{Add\:or\:subtract\:numerically\:equal\: coefficients.}}$

$\tt{\green{Step\:3:-}\orange{Solve\: obtained\:equ^n\:in\:one\:variable.}}$

$\tt{\green{Step\:4:-}\orange{Substitute\:its\:value\:in\:one\:of\:equ^n.}}$

$\tt{\green{Step\:5:-}\orange{Solve\:to\:get\:value\:of\: second\: variable.}}$

$\tt{Let's\:number\:the\:equ^ns.}$

$\blue{\tt{\leadsto \dfrac{2}{\sqrt{x}}+\dfrac{3}{\sqrt{y}}=2 .........(i)}}$

$\blue{\tt{\leadsto\dfrac{4}{\sqrt{x}}+\dfrac{9}{\sqrt{y}}=1..........(ii)}}$

$\pink{\sf{Multipling\:(i)\:by\:2.}}$

$\tt{\implies 2(\dfrac{2}{\sqrt{x}}+\dfrac{3}{\sqrt{y}})=2\times2}$

$\tt{\leadsto \dfrac{4}{\sqrt{x}}+\dfrac{6}{\sqrt{y}}=4 .........(iii)}$

_______________________________________

$\pink{\tt{Subtracting\:equ^n\:(ii)\:and\:(iii):}}$

$\tt{\implies\dfrac{4}{\sqrt{x}}+\dfrac{9}{\sqrt{y}}-( \dfrac{4}{\sqrt{x}}+\dfrac{6}{\sqrt{y}})=1-4 .}$

$\tt{\implies \dfrac{9}{\sqrt{y}}-\dfrac{6}{\sqrt{y}}=-3 .}$

$\tt{\implies \dfrac{9-6}{\sqrt{y}}=-3 .}$

$\tt{\implies \dfrac{3}{\sqrt{y}}=-3 .}$

$\tt{\implies \sqrt{y} =\dfrac{-1}{1}.}$

$\tt{\implies y = (-1)^2.}$

$\underline{\boxed{\red{\bf{\mapsto y = 1}}}}$

_________________________________________

$\pink{\tt{Putting\:this\:value\:in\:(i).}}$

$\tt{\implies \dfrac{4}{\sqrt{x}}=4-6.}$

$\tt{\implies \dfrac{4}{\sqrt{x}}=-2.}$

$\tt{\implies \sqrt{x}=\dfrac{4}{-2}}$

$\tt{\implies \sqrt{x}=(-2)}$

$\tt{\implies x=(-2)^2.}$

$\underline{\boxed{\purple{\bf{\mapsto x= 4}}}}$

$\underline{\green{\tt{\underset{\red{Required\:Answer}}{\underbrace{\hookrightarrow Hence\:value\:of\:x\:is\:4\:and\:y\:is\:1.}}}}}$

.