In the system of equations 12x+y+8x-y=8 and 27x+y-12x-y=3, the values of x and y will be

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A
$\frac{5}{3}$ and $\frac{1}{3}$
B
$\frac{5}{2}$ and $\frac{1}{2}$
C
2 and $\frac{1}{3}$
D
$\frac{5}{4}$ and $\frac{1}{4}$

Solution:

Given,

\frac{12}{x + y} + \frac{8}{x - y} = 8

and

\frac{27}{x + y} - \frac{12}{x - y} = 3

.
Let

\frac{1}{x + y} = a

and

\frac{1}{x - y} = b

.
Then the equations will be,

12 a + 8 b = 8

........ (1) and,

27 a - 12 b = 3

........... (2)
On multiplying (1) by 3 and (2) by 2 and adding.
⇒

3 (12 a + 8 b) + 2 (27 a - 12 b) = 8 \times 3 + 3 \times 2

⇒

36 a + 24 b + 54 a - 24 b = 24 + 6

⇒

90 a = 30

\Rightarrow a = \frac{30}{90}

⇒

a = \frac{1}{33}

On substituting this value of a in (1),
⇒

12 \times \frac{1}{3} + 8 b = 8

⇒

4 + 8 b = 8

⇒ 8b = 4

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

Since,

\frac{1}{x + y} = a = \frac{1}{3}

\Rightarrow x + y = 3

......(3)
⇒

\frac{1}{x + y} = b = \frac{1}{2}

\Rightarrow x - y = 2

......(4)
On adding equations (3) and (4),
⇒

2 x = 5

⇒

x = \frac{5}{2}

On substituting this value of

x

in equation (3),
⇒

\frac{5}{2} + y = 3

\Rightarrow y = 3 - \frac{5}{2}

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

∴ x = \frac{5}{2}

and

y = \frac{1}{2}

.
Hence, the correct option is 2.

In the system of equations $\frac{12}{x + y} + \frac{8}{x - y} = 8$ and $\frac{27}{x + y} - \frac{12}{x - y} = 3$ , the values of $x$ and $y$ will be

Solution:

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