Solve these equations by cross multiplication method:-

(b²/ a)* x - (a²/b)* y = ab (a + b)

b²x - a²y = 2a²b²

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Question

Solve these equations by cross multiplication method:-
(b2/ a)* x - (a2/b)* y = ab (a + b)
b2x - a2y =
2a2b2

Santosh Kumar · Accepted Answer

$(\frac{b^{2}}{a}) x - (\frac{a^{2}}{b}) y = a b (a + b) b^{2} x - a^{2} y = 2 a^{2} b^{2} (\frac{b^{2}}{a}) x - (\frac{a^{2}}{b}) y - a b (a + b) = 0 b^{2} x - a^{2} y - 2 a^{2} b^{2} = 0$

$\frac{x}{- (\frac{a^{2}}{b}) (- 2 a^{2} b^{2}) - (- a b) (a + b) (- a^{2})} = \frac{y}{(b^{2}) (- a b) (a + b) - (\frac{b^{2}}{a}) (- 2 a^{2} b^{2})} = \frac{1}{(\frac{b^{2}}{a}) (- a^{2}) - (- \frac{a^{2}}{b}) (b^{2})} \frac{x}{2 a^{4} b - a^{3} b (a + b)} = \frac{y}{- a b^{3} (a + b) + 2 a b^{4}} = \frac{1}{- a b^{2} + a^{2} b} \Rightarrow \frac{x}{2 a^{4} b - a^{4} b - a^{3} b^{2}} = \frac{y}{- a^{2} b^{3} - a b^{4} + 2 a b^{4}} = \frac{1}{a^{2} b - a b^{2}}$

$\frac{x}{a^{4} b - a^{3} b^{2}} = \frac{y}{a b^{4} - a^{2} b^{3}} = \frac{1}{a^{2} b - a b^{2}} \Rightarrow x = \frac{a^{4} b - a^{3} b^{2}}{a^{2} b - a b^{2}}$

and

$y = \frac{a b^{4} - a^{2} b^{3}}{a^{2} b - a b^{2}} \Rightarrow x = \frac{a^{3} b (a - b)}{a b (a - b)} = a^{2} and y = \frac{a b^{3} (b - a)}{a b (a - b)} = {- b}^{2} ∴ x = a^{2} and y = - b^{2}$

Solve these equations by cross multiplication method:- (b2/ a)* x - (a2/b)* y = ab (a + b) b2x - a2y = 2a2b2

Solve these equations by cross multiplication method:-

(b²/ a)* x - (a²/b)* y = ab (a + b)

b²x - a²y = 2a²b²