Q. 5 - Algebra 2

Question

Solve each system of equations.

$\begin{array}{c}9a+7b=-30\\ 8b+5c=11\\ -3a+10c=73\end{array}$

Step-by-Step Solution

Answer

The solution is $a=-1,b=-3c=7.$

Step 1. Using elimination to obtain two equations in two variables

Given equations are 

$\begin{array}{c}9a+7b=-30\\ 8b+5c=11\\ -3a+10c=73\end{array}$

Multiplying third equation by $3$ and adding to the first equation, we get

$\begin{array}{c}7b+30c=-30+219\\ 7b+30c=189\end{array}$

Step 2. Solving the system of two equations in two variables

System of equations in two variables are

$\begin{array}{c}8b+5c=11\\ 7b+30c=189\end{array}$

Multiplying the first equation by 6 and subtracting with second equation, we get

$\begin{array}{c}48b-7b=66-189\\ 41b=-123\\ b=-3\end{array}$

Putting the value of b in equation  $8b+5c=11$we get

$\begin{array}{c}8\left(-3\right)+5c=11\\ -24+5c=11\\ 5c=11+24\\ 5c=35\\ c=7\end{array}$

Step 3. Evaluating the value of a by using the values of b and c

From the equation $9a+7b=-30$ we get

$\begin{array}{c}9a+7\left(-3\right)=-30\\ 9a-21=-30\\ 9a=-30+21\\ 9a=-9\\ a=-1\end{array}$