Question

Solve the following system using Cramer’s Rule.

$5x+y=-13$

$3x-2y=0$

• A. $x=2\text{and}y=-3$
• B. $x=-4\text{and}y=-5$
• C. $x=2\text{and}y=3$
• D. $x=-2\text{and}y=-3$

Answer 1

The correct option is D $x=-2\text{and}y=-3$

$5x+y=-13$ -------------(1)

$3x-2y=0$-------------(2)

$D=\left|\begin{array}{cc}5& 1\\ 3& -2\end{array}\right|$

$=\left(5\right)\times (-2)-3\times \left(1\right)$
$=-10-3$
$=-13$

$D_x=\left|\begin{array}{cc}-13& 1\\ 0& -2\end{array}\right|$

$=(-13)\times (-2)-\left(0\right)\times \left(1\right)$
$=26-0$
$=26$

Now,
$D_y=\left|\begin{array}{cc}5& -13\\ 3& 0\end{array}\right|$

$=\left(5\right)\times \left(0\right)-\left(3\right)\times (-13)$
$=0+39$
$=39$

Again,

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

So, $x=\frac{\left(26\right)}{-13}=-2$

and

$y=\frac{\left(39\right)}{-13}=-3$

Hence,

$x=-2\text{and}y=-3$