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

$x + 2 y + 1 = 0$ ,

$2 x - 3 y - 12 = 0$ .

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Solution

$x + 2 y + 1 = 0$ ……….(i)

$2 x - 3 y - 12 = 0$ ……….. (ii)

Here $a_{1} = 1, b_{1} = 2, c_{1} = 1$

$a_{2} = 2, b_{2} = - 3, c_{2} = - 12$

By cross multiplication method,

$\frac{x}{- 24 + 3} = \frac{- y}{- 12 - 2} = \frac{1}{- 3 - 4}$

$\frac{x}{- 21} = \frac{- y}{- 14} = \frac{1}{- 7}$

Now,

$\frac{x}{- 21} = \frac{1}{- 7}$

$= x = 3$

And,

$\frac{- y}{- 14} = \frac{1}{- 7}$

$= y = - 2$

The solution of the given system of equation is 3 and -2 respectively.

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