Question

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

$$\begin{gathered} 6x+3y=6xy \\ 2x+4y=5xy \end{gathered}$$

Answer 1

$$\begin{gathered} 6x+3y=6xy \\ \Rightarrow \frac{6x+3y}{xy}=6 \\ \Rightarrow \frac{6}{y}+\frac{3}{x}=6 \end{gathered}$$

Let us substitute $\frac{1}{x}=m$ and $\frac{1}{y}=n$

$$\begin{gathered} \Rightarrow 6n+3m=6 \\ \Rightarrow 3m+6n-6=0\dots \dots \dots \dots \dots \dots \dots \dots .\left(1\right) \end{gathered}$$

$$\begin{gathered} 2x+4y=5xy \\ \Rightarrow \frac{2x+4y}{xy}=5 \\ \Rightarrow \frac{2}{y}+\frac{4}{x}=5 \end{gathered}$$

Let us substitute $\frac{1}{x}=m$ and $\frac{1}{y}=n$

$$\begin{gathered} \Rightarrow 2n+4m=5 \\ \Rightarrow 4m+2n-5=0\dots \dots \dots \dots \dots \dots \dots \dots ..\left(2\right) \\ 3m+6n–6=0 \\ 4m+2n–5=0 \end{gathered}$$

By cross-multiplication method, we get

$$\begin{gathered} \frac{m}{(-30–(-12\left)\right)}=\frac{n}{(-24-(-15\left)\right)}=\frac{1}{(6-24)} \\ \Rightarrow \frac{m}{-18}=\frac{n}{-9}=\frac{1}{-18} \\ \Rightarrow \frac{m}{-18}=\frac{1}{-18} \\ \Rightarrow m=1 \\ \text{and}\frac{n}{-9}=\frac{1}{-18} \\ \Rightarrow n=\frac{1}{2} \\ \therefore m=1\text{and}n=\frac{1}{2} \\ \Rightarrow m=\frac{1}{x}=1\text{and}n=\frac{1}{y}=\frac{1}{2} \\ \Rightarrow x=1\text{and}y=2 \end{gathered}$$

Hence, $\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$ and $\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}$ are the solution of given pair of equation.