Q16. - Algebra 2

Question

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

$2m-4n=-1$

$3n-4m=-5$

Step-by-Step Solution

Answer

The required solution is $\left(\frac{23}{10},\frac{7}{5}\right)$.

Step 1- Determine the value of m

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

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

Substitute 2 in place of a, $-4$ in place of c, $-4$ in place of b, 3 in place of d, $-1$ in place of e and $-5$ in place of f into $m=\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of m.

$\begin{array}{c}m=\frac{\left|\begin{array}{cc}-1& -4\\ -5& 3\end{array}\right|}{\left|\begin{array}{cc}2& -4\\ -4& 3\end{array}\right|}\\ =\frac{\left(-1\right)\cdot 3-\left(-4\right)\cdot \left(-5\right)}{2\cdot 3-\left(-4\right)\cdot \left(-4\right)}\\ =\frac{-23}{-10}\\ =\frac{23}{10}\end{array}$

Step 2 – Determine the value of n

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

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

Substitute 2 in place of a, $-4$ in place of c, $-4$ in place of b, 3 in place of d, $-1$ in place of e and $-5$ in place of f into $n=\frac{\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of n.

$\begin{array}{c}n=\frac{\left|\begin{array}{cc}2& -1\\ -4& -5\end{array}\right|}{\left|\begin{array}{cc}2& -4\\ -4& 3\end{array}\right|}\\ =\frac{2\cdot \left(-5\right)-\left(-1\right)\cdot \left(-4\right)}{2\cdot 3-\left(-4\right)\cdot \left(-4\right)}\\ =\frac{-14}{-10}\\ =\frac{14}{10}\end{array}$

Step 3- Write the solution

The obtained value of m is 5 and n is 1 so the solution is $\left(\frac{23}{10},\frac{7}{5}\right)$.