Question

Solve $6x+3y=6xy$ and $2x+4y=5xy$$x=1$ by reducing them to pair of linear equation.

Answer 1

Step 1. know that Any equation which can be put in the form $ax+by+c=0$, where $a,b$ and $c$ are real numbers, and a and b are not both zero, is called a linear equation in two variables.

The solution of a linear equation is not affected when:
(i) The same number is added to (or subtracted from) both the sides of the equation.
(ii) We multiply or divide both sides of the equation by the same non-zero number.

Given :

$$\begin{gathered} 6x+3y=6xy......\left(1\right) \\ 2x+4y=5xy......\left(2\right) \end{gathered}$$

Step 2: Divide equations (1) and (2) by $XY$.

$$\begin{gathered} \frac{6}{y}+\frac{3}{x}=6......\left(3\right) \\ \frac{2}{y}+\frac{4}{x}=5......\left(4\right) \end{gathered}$$

Step 3: Consider $\frac{1}{x}=m$ and $\frac{1}{y}=n$.

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

Multiply equation (6) with $3$.

$6n+12m=15......\left(7\right)$

Step 4: Subtract equation (5) from equation (7).

$\begin{array}{l}9m=9\\ \Rightarrow m=1\end{array}$

Put $m=1$in equation (5).

$\begin{array}{l}6n+3\times 1=6\\ \Rightarrow 6n=6-3\\ \Rightarrow n=\frac{3}{6}\\ \Rightarrow n=\frac{1}{2}\end{array}$

Step 5: Subtitute $m=1$in $\frac{1}{x}=m$.

$\begin{array}{l}\frac{1}{x}=1\\ \Rightarrow x=1\end{array}$

Subtitute $n=\frac{1}{2}$in $\frac{1}{y}=n$.

$\begin{array}{l}\frac{1}{y}=\frac{1}{2}\\ \Rightarrow y=2\end{array}$

Hence $x=1$and $y=2$.