Question

Solve the following pairs of linear equations:

$$\begin{gathered} \frac{x}{3}+\frac{y}{4}=4 \\ \frac{5x}{6}-\frac{y}{8}=4 \end{gathered}$$

Answer 1

Step 1: Simplify both the equations into standard form and then write one variable in terms of other.

On simplifying the equation $\frac{x}{3}+\frac{y}{4}=4$, we get

$$\begin{gathered} \frac{x}{3}+\frac{y}{4}=4 \\ \Rightarrow \frac{4x+3y}{12}=4 \\ \Rightarrow 4x+3y=48.........\left(1\right) \end{gathered}$$

and on simplifying the equation $\frac{5x}{6}-\frac{y}{8}=4$, we get

$$\begin{gathered} \frac{5x}{6}-\frac{y}{8}=4 \\ \Rightarrow \frac{20x-3y}{24}=4 \\ \Rightarrow 20x-3y=96.........\left(2\right) \end{gathered}$$

Taking equation $\left(1\right)$, we have:

$$\begin{gathered} 4x+3y=48 \\ \Rightarrow 3y=48-4x \\ \Rightarrow y=\frac{48-4x}{3}.......\left(3\right) \end{gathered}$$

Step 2: Find the value of $x$

On substituting the value of $y$ in equation $\left(2\right)$, we have

$$\begin{gathered} 20x-3\left(\frac{48-4x}{3}\right)=96 \\ \Rightarrow 20x-48+4x=96 \\ \Rightarrow 24x=96+48 \\ \Rightarrow 24x=144 \\ \Rightarrow x=\frac{144}{24} \\ \Rightarrow x=6 \end{gathered}$$

Step 3: Find the value of $y$

Now, substituting the value of $x$ in equation $\left(3\right)$, we have

$$\begin{gathered} y=\frac{48-4\left(6\right)}{3} \\ \Rightarrow y=\frac{48-24}{3} \\ \Rightarrow y=\frac{24}{3} \\ \Rightarrow y=8 \end{gathered}$$

Hence, $\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{6}$ and $\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{8}$ is the solution of given equation.