Question

Solve the following pairs of equations by reducing them to a pair of linear equations:

$\frac{10}{(x+y)}+\frac{2}{(x-y)}=4$

$\frac{15}{(x+y)}–\frac{5}{(x-y)}=-2$

Answer 1

$\frac{10}{(x+y)}+\frac{2}{(x-y)}=4$

$\frac{15}{(x+y)}–\frac{5}{(x-y)}=-2$

Let us substitute $\frac{1}{x+y}=m$ and $\frac{1}{x-y}=n$ in the given equations, we get,

$$\begin{gathered} 10m+2n=4 \\ \Rightarrow 10m+2n–4=0\dots \dots \dots \dots \dots \dots ..\dots ..\left(1\right) \\ 15m–5n=-2 \\ \Rightarrow 15m–5n+2=0\dots \dots \dots \dots \dots \dots \dots \dots ..\left(2\right) \end{gathered}$$

On cross multiplying we get,

$$\begin{gathered} \frac{m}{(4-20)}=\frac{n}{(-60-(20\left)\right)}=\frac{1}{(-50-30)} \\ \Rightarrow \frac{m}{-16}=\frac{n}{-80}=\frac{1}{-80} \\ \Rightarrow \frac{m}{-16}=\frac{1}{-80}\text{and}\frac{n}{-80}=\frac{1}{-80} \\ \Rightarrow m=\frac{1}{5}\text{and}n=1 \\ \Rightarrow m=\frac{1}{(x+y)}=\frac{1}{5} \\ \Rightarrow x+y=5\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(3\right) \\ \text{and},n=\frac{1}{(x-y)}=1 \\ \Rightarrow x-y=1\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(4\right) \end{gathered}$$

Adding equation $\left(3\right)$ and $\left(4\right)$, we get

$$\begin{gathered} 2x=6 \\ \Rightarrow x=3........\dots \dots .\left(5\right) \end{gathered}$$

Substituting value of $x=3$ in equation $\left(3\right)$, we get

$y=2$

Hence, $\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}$ and $\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}$ are the solution of given pair of equation.