Question

8. Solve the following systems of equations by 'Cross Multiplication'
method.

( a - b )x + ( a + b )y = a^2 - 2ab - b^2 ;

( a + b )(x + y ) = a^2 + b^2

Standard:- 10

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Answer 1

= > Given Equation is (a - b)x + (a + b)y = a^2 - 2ab - b^2

= > Given Equation is (a + b)(x + y) = a^2 + b^2.

We know that :

$= \ \textgreater \ \frac{x}{b1c2 - b2c1} = \frac{y}{a1c2 - a2c1} = \frac{1}{a1b2 - a2b1}$

(1)

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

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

$= \ \textgreater \ \frac{x}{(a + b)(-2b(a + b))} = \frac{1}{-2b(a + b)}$

$= \ \textgreater \ x = \frac{(a + b)(-2b(a + b))}{-2b(a + b))}$

= > x = a + b.




(2) 

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

$= \ \textgreater \ \frac{y}{-a^3 - a^2b + 2a^2b +2ab^2 +ab^2 + b^3 +a^3 + ab^2 - a^2b - b^3} = \frac{1}{a^2 - b^2 - a^2 - 2ab - b^2}$

$= \ \textgreater \ \frac{y}{4ab^2} = \frac{1}{-2b(a + b)}$

$y = \frac{-2ab}{a + b}$



Hope this helps!