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

$2 a x + 3 b y = a + 2 b$
$3 a x + 2 b y = 2 a + b$

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Solution

Given equations are:
$2 a x + 3 b y = a + 2 b$
$3 a x + 2 b y = 2 a + b$
$\Rightarrow 2 a x + 3 b y - (a + 2 b) = 0$
$\Rightarrow 3 a x + 2 b y - (2 a + b) = 0$

By cross multiplication method, we know that, for a system of linear equations in $x$ and $y$ , of the form $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ , we have:

$\frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - c_{2} a_{1}} = \frac{1}{a_{1} b_{2} - a_{2} b_{1}}$

$∴ \frac{x}{(3 b) (- (2 a + b)) - (2 b) (- (a + 2 b))} = \frac{y}{(- (a + 2 b)) (3 a) - (- (2 a + b)) (2 a)} = \frac{1}{(2 a) (2 b) - (3 a) (3 b)} \frac{x}{b^{2} - 4 a b} = \frac{y}{a^{2} - 4 a b} = \frac{1}{4 a b - 9 a b} \frac{x}{- b (4 a - b)} = \frac{y}{- a (4 b - a)} = \frac{1}{- 5 a b} \Rightarrow \frac{x}{- b (4 a - b)} = \frac{1}{- 5 a b} o r, x = \frac{4 a - b}{5 a} \Rightarrow \frac{y}{- a (4 b - a)} = \frac{1}{- 5 a b} o r, y = \frac{4 b - a}{5 b} ∴ x = \frac{4 a - b}{5 a} a n d y = \frac{4 b - a}{5 b}$

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Similar questions

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

$2 a x + 3 b y = (a + 2 b), 3 a x + 2 b y = (2 a + b)$

2 a x + 3 b y = (a + 2 b), 3 a x + 2 b y = (2 a + b) .

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

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

5 a x + 6 b y = 28

3 a x + 4 b y = 18

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