Q50. - Algebra 2

Question

Solve each system of equations.

 

 $\begin{array}{l}2a+b=2\\ 5a=15\\ a+b+c=-1\end{array}$

Step-by-Step Solution

Answer

The solution of the system of equations:  $\begin{array}{c}2a+b=2\\ 5a=15\\ a+b+c=-1\end{array}$ is $\left(3,-4,0\right).$

Step 1 - Solve for variables a

Let the equations be:

 

$2a+b=2$               … (i)

$5a=15$                 … (ii)

$a+b+c=-1$           … (iii)

Solve equation (ii) so as to obtain variable a.

$\begin{array}{c}\frac{5a}{5}=\frac{15}{5}\\ a=3\end{array}$

Step 2 - Solve for variable b

Substitute 3 for a into equation (i) so as to obtain the value of b.

$\begin{array}{c}2\left(3\right)+b=2\\ 6+b-6=2-6\\ b=-4\end{array}$

Step 3 - Solve for variable c

Now substitute 3 for a and -4 for b into the equation (iii) so as to obtain variable c. 

$\begin{array}{c}\left(3\right)+\left(-4\right)+c=-1\\ 3-4+c=-1\\ -1+c+1=-1+1\\ c=0\end{array}$

Therefore, the solution is $\left(3,-4,0\right)$.