Question

Solve the following simultaneous equation using Cramer's rule.

$$\begin{gathered} 3x-4y=10 \\ 4x+3y=5 \end{gathered}$$

Answer 1

Given:

$$\begin{gathered} 3x–4y=10 \\ 4x+3y=5 \end{gathered}$$

$$\begin{gathered} D=\left|\begin{array}{cc}3& -4\\ 4& 3\end{array}\right| \\ =9+16 \\ =25 \end{gathered}$$

And,

$$\begin{gathered} Dx=\left|\begin{array}{cc}10& -4\\ 5& 3\end{array}\right| \\ =30+20 \\ =50 \end{gathered}$$

$$\begin{gathered} Dy=\left|\begin{array}{cc}3& 10\\ 4& 5\end{array}\right| \\ =15–40 \\ =-25 \end{gathered}$$

So, by using Cramer's rule,

$$\begin{gathered} x=\frac{D_x}{D} \\ =\frac{50}{25} \\ =2 \end{gathered}$$

and,

$$\begin{gathered} y=\frac{D_y}{D} \\ =\frac{-25}{25} \\ =-1 \end{gathered}$$

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