Question

Solve the following system using Cramer’s Rule.

$2x+3y=4$

$x+y=1$

• A. $x=-1\text{and}y=2$
• B. $x=1\text{and}y=-2$
• C. $x=-1\text{and}y=2$
• D. $x=-1\text{and}y=-2$

Answer 1

The correct option is A $x=-1\text{and}y=2$

$2x+3y=4$ -------------(1)

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

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

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

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

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

Now,
$D_y=\left|\begin{array}{cc}2& 4\\ 1& 1\end{array}\right|$

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

Again,

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

So, $x=\frac{1}{-1}=-1$

and

$y=\frac{(-2)}{-1}=2$

Hence,

$x=-1\text{and}y=2$