Q17. - Algebra 2

Question

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

$4x+3y=6$

$8x-y=-9$

Step-by-Step Solution

Answer

The required solution is $\left(-\frac{3}{4},3\right)$.

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 4 in place of a, 8 in place of c, 3 in place of b, $-1$ in place of d, 6 in place of e and $-9$ 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.

$\begin{array}{c}x=\frac{\left|\begin{array}{cc}6& 3\\ -9& -1\end{array}\right|}{\left|\begin{array}{cc}4& 3\\ 8& -1\end{array}\right|}\\ =\frac{6\cdot \left(-1\right)-3\cdot \left(-9\right)}{4\cdot \left(-1\right)-3\cdot 8}\\ =\frac{21}{-28}\\ =-\frac{3}{4}\end{array}$

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 4 in place of a, 8 in place of c, 3 in place of b, $-1$ in place of d, 6 in place of e and $-9$ 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.

$\begin{array}{c}y=\frac{\left|\begin{array}{cc}4& 6\\ 8& -9\end{array}\right|}{\left|\begin{array}{cc}4& 3\\ 8& -1\end{array}\right|}\\ =\frac{4\cdot \left(-9\right)-6\cdot 8}{4\cdot \left(-1\right)-3\cdot 8}\\ =\frac{-84}{-28}\\ =3\end{array}$

Step 3- Write the solution

The obtained value of x is $-\frac{3}{4}$ and y is 3 so the solution is $\left(-\frac{3}{4},3\right)$.