Solve the equations using cross multiplication method:
$x - 3 y = 8$ and $x - 9 y = 14$

x = - 5, y = 1

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x = 5, y = 1

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x = 5, y = - 1

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x = - 5, y = - 1

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Solution

The correct option is C $x = 5, y = - 1$
$x - 3 y = 8$ ---(1)
$x - 9 y = 14$ ---(2)
Using formula for cross multiplication method:
$\frac{x}{(b_{1} c_{2} - b_{2} c_{1})} = \frac{y}{(c_{1} a_{2} - a_{1} c_{2})} = \frac{- 1}{(a_{1} b_{2} - a_{2} b_{1})}$
So, from equation (1) and (2) we can write the value of $a, b$ and $c$ .
$\frac{x}{- 3 \times 14 - (- 9) \times 8} = \frac{y}{8 \times 1 - 1 \times 14} = \frac{- 1}{1 \times (- 9) - 1 \times (- 3)}$
$\frac{x}{- 42 + 72} = \frac{y}{8 - 14} = \frac{- 1}{- 9 + 3}$
$\frac{x}{30} = \frac{y}{- 6} = \frac{- 1}{- 6}$
$\frac{x}{30} = \frac{- 1}{- 6}$
$- 6 x = - 30$
$x = 5$
$\frac{y}{- 6} = \frac{- 1}{- 6}$
$- 6 y = 6$
$y = - 1$
Therefore, $x = 5, y = - 1$

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Similar questions

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

$\frac{5}{(x + y)} - \frac{2}{(x - y)} + 1 = 0, \frac{15}{(x + y)} + \frac{7}{(x - y)} - 10 = 0. (x \neq y, x \neq - y) .$

\frac{5}{(x + y)} - \frac{2}{(x - y)} + 1 = 0 \frac{15}{(x + y)} + \frac{7}{(x - y)} - 10 = 0 (x \neq y, x \neq - y)

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

$2 x + 5 y = 1, 2 x + 3 y = 3.$

x - y = 12

x + y = 14

\frac{x}{5} + \frac{y}{15} = \frac{1}{3} \dots . (1)

\frac{x}{3} + \frac{y}{5} = \frac{1}{5} \dots (2)

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