Q5. - Algebra 2

Question

Solve each system of equations.

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

Step-by-Step Solution

Answer

 The solution is $a=-1, b=-3, c=7.$

Step-1 –Using elimination to obtain two equations in two variables

Given equations are 

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

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

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

Step-2 –Solving the system of two equations in two variables

System of equations in two variables are

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

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

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

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

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

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

From the equation $9a+7b=-30$ we get

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