Question

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

$$\begin{gathered} \frac{(7x-2y)}{xy}=5 \\ \frac{(8x+7y)}{xy}=15 \end{gathered}$$

Answer 1

Solve the following pairs of equations:

$$\begin{gathered} \frac{(7x-2y)}{xy}=5 \\ \Rightarrow \frac{7}{y}–\frac{2}{x}=5\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(1\right) \\ \frac{(8x+7y)}{xy}=15 \\ \Rightarrow \frac{8}{y}+\frac{7}{x}=15\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(2\right) \end{gathered}$$

Let us substitute $\frac{1}{x}=m$ in the given equation we get,

$$\begin{gathered} –2m+7n=5 \\ \Rightarrow -2+7n–5=0\dots \dots ..\left(3\right) \\ 7m+8n=15 \\ \Rightarrow 7m+8n–15=0..\dots \dots \left(4\right) \end{gathered}$$

By cross-multiplication method, we get,

$$\begin{gathered} \frac{m}{(-105-(-40\left)\right)}=\frac{n}{(-35-30)}=\frac{1}{(-16-49)} \\ \Rightarrow \frac{m}{(-65)}=\frac{n}{(-65)}=\frac{1}{(-65)} \\ \Rightarrow \frac{m}{-65}=\frac{1}{-65} \\ \Rightarrow m=1 \\ \text{and},\frac{n}{(-65)}=\frac{1}{(-65)} \\ \Rightarrow n=1 \\ \therefore m=1\text{and}n=1 \\ \Rightarrow m=\frac{1}{x}=1 \\ \Rightarrow n=\frac{1}{x}=1 \end{gathered}$$

Hence, $\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$ and $\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$ are the solution of given pair of equation.