Question

On solving $\frac{25}{x+y}-\frac{3}{x-y}=1,\frac{40}{x+y}+\frac{2}{x-y}=5$ we get:

• A. $x=8,y=6$
• B. $x=4,y=6$
• C. $x=6,y=4$
• D. None of these.

Answer 1

The correct option is C

$x=6,y=4$

Step 1: Use the given information/equations and apply the substitution method.

The given system of equations is as follows:

$\begin{array}{rcl}\frac{25}{x+y}-\frac{3}{x-y}& =& 1\\ \frac{40}{x+y}+\frac{2}{x-y}& =& 5\end{array}$

Substitute $m=\frac{1}{x+y}\text{and}n=\frac{1}{x-y}$into the above system of the linear equations as:

$\begin{array}{rcl}25m-3n& =& 1......\left(1\right)\\ 40m+2n& =& 5......\left(2\right)\end{array}$

Now, multiply equation (1) by $2$ to get:

$50m-6n=2...\left(3\right)$

Similarly, multiply (2) by $3$ as follows:

$120m+6n=15...\left(4\right)$

Now, on adding equations (3) and (4), we get:

$\begin{array}{rcl}50m+120m+6n-6n& =& 15+2\\ 170m& =& 17\\ 10m& =& 1\\ m& =& \frac{1}{10}\end{array}$

Substitute the value of $m$ into equation (2) as follows:

$\begin{array}{rcl}50\left(\frac{1}{10}\right)-6n& =& 2\\ 5-6n& =& 2\\ 6n& =& 3\\ n& =& \frac{1}{2}\end{array}$

Step 2: Find the value of $x$ and $y$

So, $\frac{1}{x+y}=\frac{1}{10}$and $\frac{1}{x-y}=\frac{1}{2}$. Thus, $x+y=10$ and $x-y=2$.

Now, add the above obtained two equations to get:

$\begin{array}{rcl}x+y+x-y& =& 10+2\\ 2x& =& 12\\ x& =& 6\end{array}$

Substitute the value of $x$ into $x+y=10$.

Therefore,

$\begin{array}{rcl}6+y& =& 10\\ y& =& 10-6\\ y& =& 4\end{array}$

So, the required value is $x=6,y=4$. Hence, option C is correct.