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Solve the following pair of linear equation by cross - multiplication method:

$2 x + 3 y = 13; 5 x − 4 y = − 2$

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x = \frac{1}{2} and y = \frac{1}{3}

x = \frac{- 1}{4} and y = \frac{5}{2}

x = \frac{- 1}{4} and y = \frac{- 5}{2}

x = \frac{- 1}{4} and y = \frac{5}{4}

answer is A.

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Detailed Solution

Given pair of linear equations are,

2 x + 3 y = 13 5 x − 4 y = − 2

The following equations should be written in standard form,

a 1 x + b 1 y + c 1 = 0

and

a 2 x + b 2 y + c 2 = 0.

The following equations are stated in standard form,

2 x + 3 y − 13 = 0...... (1) 5 x − 4 y + 2 = 0...... (2)

Putting

\frac{1}{x} = u and \frac{1}{y} = v

Then the equations become,

2 u + 3 v - 13 = 0 … … . 3 5 u - 4 v + 2 = 0 … … 4

Comparing equation (1) and equation (2) with the standard form,

a 1 = 2, b 1 = 3, c 1 = − 13 a 2 = 5, b 2 = - 4, c 2 = 2

Using cross - multiplication method, we get:
Question Image

Then,

u 32 − − 4 − 13 = v − 13 5 − 22 = 1 2 × − 4 − 5 × 3

u 6 − 52 = v − 65 − 4 = 1 − 8 − 15

u − 46 = v − 69 = 1 − 23

I         II        III

On taking I and III terms:

u − 46 = 1 − 23 − 23 u = − 46 u = 4623 u = 2

On taking II and III terms:

v − 69 = 1 − 23 − 23 v = − 69 v = 6923 v = 3

On putting the values of u and v in

\frac{1}{x} = u and \frac{1}{y} = v

we get,

\frac{1}{x} = u = 2

\Rightarrow x = \frac{1}{2}

and

\frac{1}{y} = v = 3

\Rightarrow y = \frac{1}{3}

So, the required solutions are

x = \frac{1}{2} and y = \frac{1}{3}

.
Hence, option (1) is correct.

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