Q20. - Algebra 2

Question

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

$3x-2y=4$

$\frac{1}{2}x-\frac{2}{3}y=1$

Step-by-Step Solution

Answer

The required solution is$\left(\frac{2}{3},-1\right)$.

Step 1- Determine the value of x

Apply the Cramer’s rule for the value of x. 

$x=\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$

Substitute 3 in place of a, $\frac{1}{2}$ in place of c, $-2$ in place of b, $-\frac{2}{3}$ in place of d, 4 in place of e and 1 in place of f into $x=\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of x.

$\begin{array}{c}x=\frac{\left|\begin{array}{cc}4& -2\\ 1& -\frac{2}{3}\end{array}\right|}{\left|\begin{array}{cc}3& -2\\ \frac{1}{2}& -\frac{2}{3}\end{array}\right|}\\ =\frac{4\cdot \left(-\frac{2}{3}\right)-\left(-2\right)\cdot 1}{3\cdot \left(-\frac{2}{3}\right)-\left(-2\right)\cdot \frac{1}{2}}\\ =\frac{-\frac{2}{3}}{-1}\\ =\frac{2}{3}\end{array}$

Step 2 – Determine the value of y

Apply the Cramer’s rule for the value of y. 

$y=\frac{\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$

Substitute 3 in place of a, $\frac{1}{2}$ in place of c, $-2$ in place of b, $-\frac{2}{3}$ in place of d, 4 in place of e and 1 in place of f into $y=\frac{\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of y.

$\begin{array}{c}y=\frac{\left|\begin{array}{cc}3& 4\\ \frac{1}{2}& 1\end{array}\right|}{\left|\begin{array}{cc}3& -2\\ \frac{1}{2}& -\frac{2}{3}\end{array}\right|}\\ =\frac{3\cdot 1-4\cdot \left(\frac{1}{2}\right)}{3\cdot \left(-\frac{2}{3}\right)-\left(-2\right)\cdot \frac{1}{2}}\\ =\frac{1}{-1}\\ =-1\end{array}$

Step 3- Write the solution

The obtained value of x is $\frac{2}{3}$ and y is $-1$ so the solution is $\left(\frac{2}{3},-1\right)$.