Q5. - Algebra 2

Question

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

$0.2a=0.3b$

$0.4a-0.2b=0.2$

Step-by-Step Solution

Answer

The required solution is $\left(0.75,0.5\right)$.

Step 1- Determine the value of a

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

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

Substitute $0.2$ in place of a, $0.4$ in place of c, $-0.3$ in place of b, $-0.2$ in place of d, 0 in place of e and $0.2$ in place of f into $a=\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of a.

$\begin{array}{c}a=\frac{\left|\begin{array}{cc}0& -0.3\\ 0.2& -0.2\end{array}\right|}{\left|\begin{array}{cc}0.2& -0.3\\ 0.4& -0.2\end{array}\right|}\\ =\frac{0\cdot \left(-0.2\right)-\left(-0.3\right)\cdot 0.2}{0.2\cdot \left(-0.2\right)-\left(-0.3\right)\cdot 0.4}\\ =\frac{0.06}{0.08}\\ =0.75\end{array}$

Step 2 – Determine the value of b

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

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

Substitute $0.2$ in place of a, $0.4$ in place of c, $-0.3$ in place of b, $-0.2$ in place of d, 0 in place of e and $0.2$ in place of f into $b=\frac{\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of b.

$\begin{array}{c}b=\frac{\left|\begin{array}{cc}0.2& 0\\ 0.4& 0.2\end{array}\right|}{\left|\begin{array}{cc}0.2& -0.3\\ 0.4& -0.2\end{array}\right|}\\ =\frac{\left(0.2\right)\cdot \left(0.2\right)-0\cdot \left(0.4\right)}{0.2\cdot \left(-0.2\right)-\left(-0.3\right)\cdot 0.4}\\ =\frac{0.04}{0.08}\\ =0.5\end{array}$

Step 3- Write the solution

The obtained value of a is $0.75$ and b is $0.5$ so the solution is $\left(0.75,0.5\right)$.