Q15. - Algebra 2

Question

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

$3a+5b=33$

$5a+7b=51$

Step-by-Step Solution

Answer

The required solution is $\left(6,3\right)$.

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 3 in place of a, 5 in place of c, 5 in place of b, 7 in place of d, 33 in place of e and 51 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.

$\begin{array}{c}a=\frac{\left|\begin{array}{cc}33& 5\\ 51& 7\end{array}\right|}{\left|\begin{array}{cc}3& 5\\ 5& 7\end{array}\right|}\\ =\frac{33\cdot 7-5\cdot 51}{3\cdot 7-5\cdot 5}\\ =\frac{-24}{-4}\\ =6\end{array}$

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 3 in place of a, 5 in place of c, 5 in place of b, 7 in place of d, 33 in place of e and 51 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.

$\begin{array}{c}b=\frac{\left|\begin{array}{cc}3& 33\\ 5& 51\end{array}\right|}{\left|\begin{array}{cc}3& 5\\ 5& 7\end{array}\right|}\\ =\frac{3\cdot 51-33\cdot 5}{3\cdot 7-5\cdot 5}\\ =\frac{-12}{-4}\\ =3\end{array}$

Step 3- Write the solution

The obtained value of a is 6 and b is 3 so the solution is $\left(6,3\right)$.