Question

please provide step by step answer class 10 maths​

Answer 1

Answer:

x = a^2 , y = b^2

Step-by-step explanation:

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Answer 2

Given :

The given system of linear equation -

⠀⠀⠀⌬ $\sf\:\frac{x}{a}+\frac{y}{b}=a+b$

⠀⠀⠀⌬ $\sf\:\frac{x}{a^{2}}+\frac{y}{b^{2}}=2$

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To find :

• Value of x and y.

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Assumption :

Let -

⠀⠀⠀⌬ $\sf\:\frac{x}{a}+\frac{y}{b}=a+b$⠀⠀⠀ $\small{\mathfrak\gray{[~equ^{n}~1~]}}$

⠀⠀⠀⌬ $\sf\:\frac{x}{a^{2}}+\frac{y}{b^{2}}=2$⠀⠀⠀⠀ $\small{\mathfrak\gray{[~equ^{n}~2~]}}$

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Solution :

Taking equation 1 and multiplying it by 1/a , we get -

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$\large\sf\:(\frac{x}{a}+\frac{y}{b}=a+b)$ $\bf\:\times \frac{1}{a}$

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$\large\sf\implies\:\frac{x}{a} \times \frac{1}{a}+\frac{y}{b} \times \frac{1}{a}=(a+b) \times \frac{1}{a}$

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$\large\sf\implies\:\frac{x}{a^{2}}+\frac{y}{ab}=\frac{a+b}{a}$

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Let -

⠀⠀⠀⌬ $\sf\:\frac{x}{a^{2}}+\frac{y}{ab}=\frac{a+b}{a}$⠀⠀⠀ $\small{\mathfrak\gray{[~equ^{n}~3~]}}$

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Subtracting equation 2 and equation 3 , we get -

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$\large\sf\implies\:(\frac{x}{a^{2}}+\frac{y}{b^{2}})-(\frac{x}{a^{2}}+\frac{y}{ab})=2-\frac{a+b}{a}$

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$\large\sf\implies\:\frac{x}{a^{2}}+\frac{y}{b^{2}}-\frac{x}{a^{2}}-\frac{y}{ab}=2-\frac{a+b}{a}$

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$\large\sf\implies\:\frac{x}{a^{2}}-\frac{x}{a^{2}}+\frac{y}{b^{2}}-\frac{y}{ab}=2-\frac{a+b}{a}$

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$\large\sf\implies\:\cancel{\frac{x}{a^{2}}}-\cancel{\frac{x}{a^{2}}}+\frac{y}{b^{2}}-\frac{y}{ab}=2-\frac{a+b}{a}$

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$\large\sf\implies\:\frac{y}{b^{2}}-\frac{y}{ab}=2-\frac{a+b}{a}$

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$\large\sf\implies\:\frac{y}{b}(\frac{1}{b}-\frac{1}{a})=\frac{2a-a-b}{a}$

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$\large\sf\implies\:\frac{y}{b}(\frac{a-b}{ab})=\frac{a-b}{a}$

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$\large\sf\implies\:\frac{y}{b}=\frac{\frac{a-b}{a}}{\frac{a-b}{ab}}$

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$\large\sf\implies\:\frac{y}{b}=\frac{a-b}{a} \times \frac{ab}{a-b}$

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$\large\sf\implies\:\frac{y}{b}=\frac{\cancel{a-b}}{a} \times \frac{ab}{\cancel{a-b}}$

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$\large\sf\implies\:\frac{y}{b}=\frac{1}{a} \times \frac{ab}{1}$

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$\large\sf\implies\:\frac{y}{b}=\frac{ab}{a}$

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$\large\sf\implies\:\frac{y}{b}=\frac{\cancel{a}b}{\cancel{a}}$

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$\large\sf\implies\:\frac{y}{b}=\frac{b}{1}$

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$\small{\underline{\boxed{\mathrm{\implies\:y~=~b^{2}}}}}$ $\bf\color{red}{⋆}$

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Substituting the value of y in equation 2 , we get -

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$\large\sf\:\frac{x}{a^{2}}+\frac{y}{b^{2}}=2$

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$\large\sf\implies\:\frac{x}{a^{2}}+\frac{b^{2}}{b^{2}}=2$

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$\large\sf\implies\:\frac{x}{a^{2}}+\cancel{\frac{b^{2}}{b^{2}}}=2$

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$\large\sf\implies\:\frac{x}{a^{2}}+1=2$

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$\large\sf\implies\:\frac{x+a^{2}}{a^{2}}=2$

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$\small\sf\implies\:x+a^{2}=2 \times a^{2}$

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$\small\sf\implies\:x+a^{2}=2a^{2}$

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$\small\sf\implies\:x=2a^{2}-a^{2}$

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$\small{\underline{\boxed{\mathrm{\implies\:x~=~a^{2}}}}}$ $\bf\color{red}{⋆}$

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$\bf\therefore~The~value~of~x~is~a^{2}~and~y~is~b^{2}$.

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Verification :

Plugging the value of x and y in equation 1 -

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$\large\sf\:\frac{x}{a}+\frac{y}{b}=a+b$

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$\large\sf\leadsto\:\frac{a^{2}}{a}+\frac{b^{2}}{b}=a+b$

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$\large\sf\leadsto\:\frac{a^{2}b+ab^{2}}{ab}=a+b$

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$\small\sf\leadsto\:a^{2}b+ab^{2}=ab(a+b)$

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$\small\sf\leadsto\:a^{2}b+ab^{2}=a^{2}b+ab^{2}$

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$\bf\leadsto\:LHS~=~RHS$

Hence Verified ~

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