Question

solve the pair of linear equations. x/a+y/b=a+b and x/a^2+y/b^2=2 (a,b not equal to 0)​

Answer 1

Step-by-step explanation:

Given:

$\frac{x}{a} + \frac{y}{b} = a + b \: \: \: ...eq1 \\ \\ \frac{x}{ {a}^{2} } + \frac{y}{ {b}^{2} } = 2 \: \: \: ...eq2$

To find: Value of x and y.

Solution:

Step 1: Multiply eq1 by 1/a and subtract both equations

$\frac{x}{ {a}^{2} } + \frac{y}{ab} = \frac{1}{a} (a + b) \\ \\ or \\ \\ \frac{x}{ {a}^{2} } + \frac{y}{ab} = 1 + \frac{b}{a} \\ \\ \frac{x}{ {a}^{2} } + \frac{y}{ {b}^{2} } = 2 \\ ( - ) \: \: \: \: ( - ) \: \: \: ( - ) \\ - - - - - - - \\ \frac{y}{ab} - \frac{y}{ {b}^{2} } = 1 + \frac{b}{a} - 2$

Take y/b common

$\frac{y}{b}\left ( \frac{1}{a} - \frac{1}{b}\right ) = \frac{b}{a} - 1$

Take LCM and solve

$\frac{y}{b} \left( \frac{b - a}{ab} \right) = \frac{b - a}{a} \\ \\ \frac{y}{b} \left( \frac{ \cancel{ \red{b - a}}}{ \cancel \red{a}b}\right ) = \frac{ \cancel{\red{b - a}}}{\cancel \red{a}} \\ \\ \frac{y}{b}\left ( \frac{1}{b}\right ) = 1$

cross multiply,remain y in LHS

$\frac{y}{ {b}^{2} } = 1 \\ \\ \bold{\green{y = {b}^{2}}}$

Step 2: Put the value of y in eq2

$\frac{x}{ {a}^{2} } + \frac{ {b}^{2} }{ {b}^{2} } = 2 \\ \\ \frac{x}{ {a}^{2} } + 1 = 2 \\ \\ \frac{x}{ {a}^{2} } = 2 - 1 \\ \\ \frac{x}{ {a}^{2} } = 1 \\ \\\bold{\green{ x = {a}^{2} }}$

Final answer:

$\bold{\red{x = {a}^{2}}} \\ \\ \bold{\red{y = {b}^{2} }}$

Hope it helps you.

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