Q8. - Algebra 2

Question

Use Cramer’s rule to solve each system of equations.

8.

$\begin{array}{l}\mathbf{3}\mathbf{x}-\mathbf{2}\mathbf{y}=\mathbf{7}\\ \mathbf{4}\mathbf{x}-\mathbf{y}=\mathbf{6}\end{array}$

Step-by-Step Solution

Answer

The solution of the given system of equations is $\left(\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{2}\right)$.

Step 1 ­- Description of step.

The solution of the system of linear equations in two variables:

$\begin{array}{l}{a}_{1}x+b_{1}y=c_1\\ a_{2}x+b_{2}y=c_2\end{array}$

by Cramer’s rule is given by $\left(x,y\right)$ where$x=\frac{\left|\begin{array}{cc}{c}_1& b_1\\ c_2& b_2\end{array}\right|}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}$ $y=\frac{\left|\begin{array}{cc}{a}_1& c_1\\ a_2& c_2\end{array}\right|}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}$and$\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|\ne 0$

Step 2 ­- Description of step.

The given system of linear equations in two variables is:

$\begin{array}{l}3x-2y=7\\ 4x-y=6\end{array}$

Therefore, by comparing the given system of linear equations with the system of linear equations $\begin{array}{l}{a}_{1}x+b_{1}y=c_1\\ a_{2}x+b_{2}y=c_2\end{array}$it can be obtained that:

$a_1=3$, $b_1=-2$, $c_1=7$, $a_2=4$, $b_2=-1$ and $c_2=6$.

Step 3 ­- Find the values of x and y.

The value of $x$ is given by:

$\begin{array}{c}x=\frac{\left|\begin{array}{cc}7& -2\\ 6& -1\end{array}\right|}{\left|\begin{array}{cc}3& -2\\ 4& -1\end{array}\right|}\\ =\frac{-7-\left(-12\right)}{-3-\left(-8\right)}\\ =\frac{-7+12}{-3+8}\\ =\frac{5}{5}\\ =1\end{array}$

The value of $y$is given by:

$\begin{array}{c}y=\frac{\left|\begin{array}{cc}3& 7\\ 4& 6\end{array}\right|}{\left|\begin{array}{cc}3& -2\\ 4& -1\end{array}\right|}\\ =\frac{18-\left(28\right)}{-3-\left(-8\right)}\\ =\frac{18-28}{-3+8}\\ =\frac{-10}{5}\\ =-2\end{array}$

The values of $x$ and \[y\] are 1 and $-2$ respectively.

Step 4 ­- Description of step.

The solution of the given system of equations is $\left(1,-2\right)$.