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Solve the following simultaneous equations: $\frac{1}{3 x} + \frac{1}{5 y} = \frac{1}{15}; \frac{1}{2 x} + \frac{1}{3 y} = \frac{1}{12}$ .

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x = 5, y = 3

x = 6, y = 6

x = 2, y = - 2

x = 1, y = - 5

answer is C.

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Detailed Solution

Given that,

\frac{1}{3 x} + \frac{1}{5 y} = \frac{1}{15}; \frac{1}{2 x} + \frac{1}{3 y} = \frac{1}{12}

.
Substituting

\frac{1}{x} = X

and

\frac{1}{y} = Y

in given equations, we get,

\frac{X}{3} + \frac{Y}{5} = \frac{1}{15}

\Rightarrow 5 X + 3 Y = 1

……..(1)

\frac{X}{2} + \frac{Y}{3} = \frac{1}{12}

\Rightarrow 6 X + 4 Y = 1

………(2)
Multiplying (1) by 6 and (2) by 5 and subtracting, we get,

30 X + 18 Y - 30 X - 20 Y = 6 - 5

- 2 Y = 1

∴ Y = - \frac{1}{2}

Substituting

Y = - \frac{1}{2}

in (1), we get,

5 X + 3 \times - \frac{1}{2} = 1

\Rightarrow 5 X = \frac{5}{2}

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

∴ x = \frac{1}{X} = 2, y = \frac{1}{Y} = - 2

(

\frac{1}{x} = X

and

\frac{1}{y} = Y

)
Therefore, option (3) is correct.

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