Question

Solve the given equations simultaneously. The equations are:

$$\begin{gathered} 4m+6n=54 \\ 3m+2n=28 \end{gathered}$$

Answer 1

$$\begin{gathered} 4m+6n=54......\left(1\right) \\ 3m+2n=28......\left(2\right) \end{gathered}$$

Multiply equation $\left(2\right)$ by $3$, we obtain

$9m+6n=84.....\left(3\right)$

Subtracting equation $\left(1\right)$ from equation $\left(3\right)$, we obtain

$$\begin{gathered} 9m+6n-4m-6n=84-54 \\ \Rightarrow 5m=30 \\ \Rightarrow m=\frac{30}{5} \\ \Rightarrow m=6 \end{gathered}$$

Substituting the value of $m$ in equation $\left(2\right)$, we obtain

$$\begin{gathered} 3\left(6\right)+2n=28 \\ \Rightarrow 18+2n=28 \\ \Rightarrow 2n=28-18 \\ \Rightarrow 2n=10 \\ \Rightarrow n=\frac{10}{2} \\ \Rightarrow n=5 \end{gathered}$$

Hence, $\mathit{m}\mathbf{}\mathbf{=}\mathbf{}\mathbf{6}$ and $\mathit{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}$ are the solution of given pair of equations.