Question

Solve the following system using Cramer’s Rule.

$-2x+4y=1$

$-2x+y=-3$

• A. $x=\frac{13}{6}\text{and}y=\frac{8}{3}$
• B. $x=\frac{13}{6}\text{and}y=\frac{4}{3}$
• C. $x=\frac{13}{6}\text{and}y=\frac{-2}{3}$
• D. $x=\frac{13}{6}\text{and}y=\frac{2}{3}$

Answer 1

The correct option is B $x=\frac{13}{6}\text{and}y=\frac{4}{3}$

$-2x+4y=1$ -------------(1)

$-2x+y=-3$-------------(2)

$D=\left|\begin{array}{cc}-2& 4\\ -2& 1\end{array}\right|$

$=(-2)\times \left(1\right)-(-2)\times \left(4\right)$
$=-2+8$
$=6$

$D_x=\left|\begin{array}{cc}1& 4\\ -3& 1\end{array}\right|$

$=\left(1\right)\times \left(1\right)-(-3)\times \left(4\right)$
$=1+12$
$=13$

Now,
$D_y=\left|\begin{array}{cc}-2& 1\\ -2& -3\end{array}\right|$

$=(-2)\times (-3)-(-2)\times \left(1\right)$
$=6-(-2)$
$=6+2$
$=8$

Again,

$x=\frac{\left(D_x\right)}{D}$ and $y=\frac{\left(D_y\right)}{D}$

So, $x=\frac{13}{6}$

and

$y=\frac{8}{6}=\frac{4}{3}$

Hence,

$x=\frac{13}{6}\text{and}y=\frac{4}{3}$