Question

Solve the following simultaneous equations using Cramer's rule.

$3x+y=1;2x=11y+3$

• A. $x=\frac{2}{5},y=\frac{-7}{3}$
• B. $x=\frac{2}{5},y=\frac{-1}{5}$
• C. $x=\frac{2}{5},y=\frac{1}{5}$
• D. $x=\frac{2}{5},y=\frac{7}{5}$

Answer 1

The correct option is B $x=\frac{2}{5},y=\frac{-1}{5}$

Given equations are $3x+y=1$ and $2x=11y+3$

For cramers rule, We first find

$D=\left(\begin{array}{cc}3& 1\\ 2& -11\end{array}\right)$

$\Rightarrow \left|D\right|=\left|\begin{array}{cc}3& 1\\ 2& -11\end{array}\right|=3\times -11-1\times 2=-33-2=-35$

For $x$ matrix

$D_x=\left[\begin{array}{cc}1& 1\\ 3& -11\end{array}\right]\Rightarrow \left[D_x\right]=\left|\begin{array}{cc}1& 1\\ 3& -11\end{array}\right|=1\times -11-1\times 3=-11-3=-14$

For $y$ matrix

$D_y=\left[\begin{array}{cc}3& 1\\ 2& 3\end{array}\right]\Rightarrow \left[D_y\right]=\left|\begin{array}{cc}3& 1\\ 2& 3\end{array}\right|=3\times 3-1\times 2=9-2=7$

Therefore, $x=\frac{D_x}{D}=\frac{-14}{-35}=\frac{2}{5}$

and $y=\frac{D_y}{D}=\frac{7}{-35}=-\frac{1}{5}$

Answer is option B.