Solve the following systems of equations by the method of cross-multiplication:

$b x + c y = a + b$

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

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Solution

Given equations:-
$b x + c y = a + b \Rightarrow b x + c y - (a + b) = 0 a x (\frac{1}{a - b} - \frac{1}{a + b}) + c y (\frac{1}{b - a} - \frac{1}{b + a}) = \frac{2 a}{a + b} \Rightarrow a x (\frac{a + b - a + b}{a^{2} - b^{2}}) + c y (\frac{b + a - b + a}{b^{2} - a^{2}}) = \frac{2 a}{a + b} \Rightarrow \frac{2 a b x}{a^{2} - b^{2}} + \frac{2 a c y}{b^{2} - a^{2}} = \frac{2 a}{a + b} \Rightarrow \frac{2 a}{a + b} (\frac{b x}{a - b} + \frac{c y}{b - a}) = \frac{2 a}{a + b} \Rightarrow \frac{1}{a - b} (b x - c y) = \frac{2 a}{a + b} \times \frac{a + b}{2 a} \Rightarrow b x - c y = a - b \Rightarrow b x - c y - (a - b) = 0$

Now, the two equations are:
$b x + c y - (a + b) = 0$
$b x - c y - (a - b) = 0$
Using cross-multiplication method,
$\frac{x}{(c) (- (a - b)) - (- c) (- (a + b))} = \frac{y}{(- (a + b)) (b) - (- (a - b)) (b)} = \frac{1}{(b) (- c) - (b) (c)} \frac{x}{- a c + b c - a c - b c} = \frac{y}{- a b - b^{2} + a b - b^{2}} = \frac{1}{- b c - b c} \frac{x}{- 2 a c} = \frac{y}{- 2 b^{2}} = \frac{1}{- 2 b c} \Rightarrow \frac{x}{- 2 a c} = \frac{1}{- 2 b c} o r, x = \frac{a}{b} \Rightarrow \frac{y}{- 2 b^{2}} = \frac{1}{- 2 b c} o r, y = \frac{b}{c} ∴ x = \frac{a}{b} a n d y = \frac{b}{c}$

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Similar questions

Solve each of the following systems of equations by the method of cross-multiplication:

$b x + c y = a + b$

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

b x + c y = a + b a x (\frac{1}{a - b} - \frac{1}{a + b}) + c y (\frac{1}{b - a} - \frac{1}{b + a}) = \frac{2 a}{a + b}

a x + b y = a - b

b x - a y = a + b

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

(a - 2 b) x + (2 a + b) y = 3

\frac{a x}{b} - \frac{b y}{a} = a + b

a x - b y = 2 a b

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