Q55. - Algebra 2

Question

Solve each system by using substitution or elimination.

 

 $\begin{array}{c}a+5b=1\\ 7a-2b=44\end{array}$

Step-by-Step Solution

Answer

The solution of the system of equations:  $\begin{array}{c}a+5b=1\\ 7a-2b=44\end{array}$ is $\left(6,-1\right)$. 

Step 1 - Solve the first equation for a in terms of b

Write the first equation $a+5b=1$ for a in terms of b.

 

$\begin{array}{c}a+5b=1\\ a+5b-5b=1-5b\\ a=1-5b\end{array}$

Step 2 - Solve for b

Substitute $1-5b$ for a into equation $7a-2b=44$ so as to obtain the value of b.

 

$\begin{array}{c}7\left(1-5b\right)-2b=44\\ 7\left(1\right)-7\left(5b\right)-2b=44\\ 7-35b-2b=44\\ 7-37b-7=44-7\\ -37b=37\\ b=-1\end{array}$

Step 3 - Solve for a

Now substitute -1 for b into either original equation and solve for a. 

$\begin{array}{c}a+5\left(-1\right)=1\\ a-5+5=1+5\\ a=6\end{array}$

Therefore, the solution of the system is $\left(6,-1\right)$.