Q18. - Algebra 2

Question

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

$0.5r-s=-1$

$0.75r+0.5s=-0.25$

Step-by-Step Solution

Answer

The required solution is $\left(-0.75,0.625\right)$.

Step 1- Determine the value of r

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

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

Substitute $0.5$ in place of a, $0.75$ in place of c, $-1$ in place of b, $0.5$ in place of d, $-1$ in place of e and $-0.25$ in place of f into $r=\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of r.

$\begin{array}{c}r=\frac{\left|\begin{array}{cc}-1& -1\\ -0.25& 0.5\end{array}\right|}{\left|\begin{array}{cc}0.5& -1\\ 0.75& 0.5\end{array}\right|}\\ =\frac{\left(-1\right)\cdot 0.5-\left(-0.25\right)\cdot \left(-1\right)}{0.5\cdot 0.5-0.75\cdot \left(-1\right)}\\ =\frac{-0.75}{1}\\ =-0.75\end{array}$

Step 2 – Determine the value of s

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

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

Substitute $0.5$ in place of a, $0.75$ in place of c, $-1$ in place of b, $0.5$ in place of d, $-1$ in place of e and $-0.25$ in place of f into $s=\frac{\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of s.

$\begin{array}{c}s=\frac{\left|\begin{array}{cc}0.5& -1\\ 0.75& -0.25\end{array}\right|}{\left|\begin{array}{cc}0.5& -1\\ 0.75& 0.5\end{array}\right|}\\ =\frac{0.5\cdot \left(-0.25\right)-0.75\cdot \left(-1\right)}{0.5\cdot 0.5-0.75\cdot \left(-1\right)}\\ =\frac{0.625}{1}\\ =0.625\end{array}$

Step 3- Write the solution

The obtained value of r is $-0.75$ and s is $0.625$ so the solution is $\left(-0.75,0.625\right)$.