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The values of $x$ and $y$ of $x + y = 15$ and $x - y = 3$ by the method of cross multiplication are respectively

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9, - 6

- 9, 6

9, 6

- 9, - 6

answer is C.

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Detailed Solution

For the pair of linear equations

a_{1} x + b_{1} y + c_{1} = 0

and

a_{2} x + b_{2} y + c_{2} = 0

, then the values of

x

and

y

can be find out by:

\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}}

For linear equations

x + y = 15

and

x - y = 3

x + y - 15 = 0

and

x - y - 3 = 0

, we have

a_{1} = 1, b_{1} = 1

and

c_{1} = - 15

and

a_{2} = 1

b_{2} = - 1

and

c_{2} = - 3

By cross multiplication method,

\frac{x}{(1) (- 3) - (- 1) (- 15)} = \frac{y}{(- 15) (1) - (- 3) (1)} = \frac{1}{(1) (- 1) - (1) (1)}

\frac{x}{- 3 - 15} = \frac{y}{- 15 + 3} = \frac{1}{- 1 - 1}

\frac{x}{- 18} = \frac{y}{- 12} = \frac{1}{- 2}

On comparing, we get

\frac{x}{- 18} = \frac{1}{- 2}

and

\frac{y}{- 12} = \frac{1}{- 2}

⇒

x = 9

and

y = 6

Therefore, the correct option is (3).

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