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Solve the following simultaneous equations: $\frac{1}{x} + \frac{1}{y} = 8; \frac{4}{x} - \frac{2}{y} = 2$ .

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x = \frac{1}{3}, y = \frac{1}{5}

x = \frac{1}{2}, y = \frac{1}{5}

x = \frac{1}{3}, y = \frac{1}{7}

x = \frac{1}{2}, y = \frac{1}{7}

answer is A.

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Detailed Solution

Given equations are

\frac{1}{x} + \frac{1}{y} = 8; \frac{4}{x} - \frac{2}{y} = 2

.
Let

\frac{1}{x} = a, \frac{1}{y} = b

.
So,

a + b = 8

.......

(1)

4 a - 2 b = 2

......

(2)

Multiply equation

(1)

2

, we get,

2 a + 2 b = 16

........

(3)

Adding equations

(2)

and

(3)

, we get,

4 a - 2 b + 2 a + 2 b = 2 + 16

\Rightarrow 6 a = 18

\Rightarrow a = \frac{18}{6}

\Rightarrow a = 3

Substituting

a = 3

a + b = 8

we get,

3 + b = 8

\Rightarrow b = 5

Now, substituting

a = 3

\frac{1}{x} = a

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

⇒

x = \frac{1}{3}

And, substituting

b = 5

\frac{1}{y} = b

\Rightarrow \frac{1}{y} = 5

⇒

y = \frac{1}{5}

Thus, the solution is

x = \frac{1}{3}

and

= \frac{1}{5}

.
Therefore, option (1) is correct.

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