Question

Find the values of $x$ and $y$ if :

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

$\frac{6}{2+x}-\frac{3}{y-4}=1$

Here, x $\ne$ -2 and y $\ne$ 4.

• A. x = -2, y = 2
• B. x = 1, y = 7
• C. x = 0, y = 8
• D. x = 7, y = -8

Answer 1

Answer: (B)

x = 1, y = 7

Let $p=\frac{1}{2+x}$ and $q=\frac{1}{y-4}$

Thus, 5p + q = 2 .... (i)

6p - 3q = 1 .... (ii)

On multiplying equation (i) by 3, we get

15p + 3q = 6 .... (iii)

On adding equations (iii) and (ii), we get

21p = 7

$\Rightarrow p=\frac{1}{3}$

On substituting the value of p in equation (i), we get

$5\times \frac{1}{3}+q=2$

$\Rightarrow q=2-\frac{5}{3}$

$\Rightarrow q=\frac{1}{3}$

Now, $p=\frac{1}{2+x}$

$\Rightarrow \frac{1}{3}=\frac{1}{2+x}$

$\Rightarrow x=1$

Similarly $q=\frac{1}{y-4}$

$\Rightarrow \frac{1}{3}=\frac{1}{y-4}$

$\Rightarrow y=7$

$\therefore$ x = 1 and y = 7