Question

Solve the following system using Cramer’s Rule.

$-4x+3y=-4$

$-8x+5y=0$

• A. $x=-5\text{and}y=-8$
• B. $x=8\text{and}y=-8$
• C. $x=-5\text{and}y=8$
• D. $x=5\text{and}y=8$

Answer 1

The correct option is A $x=-5\text{and}y=-8$

$-4x+3y=-4$ -------------(1)

$-8x+5y=0$-------------(2)

$D=\left|\begin{array}{cc}-4& 3\\ -8& 5\end{array}\right|$

$=(-4)\times \left(5\right)-3\times (-8)$
$=-20-(-24)$
$=-20+24$
$=4$

$D_x=\left|\begin{array}{cc}-4& 3\\ 0& 5\end{array}\right|$

$=(-4)\times \left(5\right)-\left(3\right)\times \left(0\right)$
$=-20-0$
$=-20$

Now,
$D_y=\left|\begin{array}{cc}-4& -4\\ -8& 0\end{array}\right|$

$=(-4)\times \left(0\right)-(-4)\times (-8)$
$=0-\left(32\right)$
$=-32$

Again,

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

So, $x=\frac{(-20)}{4}=-5$

and

$y=\frac{(-32)}{4}=-8$

Hence,

$x=-5\text{and}y=-8$