Q27. - Algebra 2

Question

Solve each system of equations.

$\begin{array}{c}2a+b-c=5\\ a-b+3c=9\\ 3a-6c=6\end{array}$

Step-by-Step Solution

Answer

The solution is $a=4,b=-2,c=1$.

Step-1 – Use elimination to make a system of two equations in two variables.

Given equations are-:

$\begin{array}{c}2a+b-c=\mathrm{5.......}\left(1\right)\\ a-b+3c=\mathrm{9.......}\left(2\right)\\ 3a-6c=\mathrm{6.......}\left(3\right)\end{array}$

Adding first and second equation, we get

$\begin{array}{c}2a+a-c+3c=5+9\\ 3a+2c=\mathrm{14......}\left(4\right)\end{array}$

Step-2 – Solve the system of two equations.

$\begin{array}{c}3a-6c=\mathrm{6.......}\left(3\right)\\ 3a+2c=\mathrm{14........}\left(4\right)\end{array}$

Subtracting fourth equation from third equation, we get

$\begin{array}{c}-6c-2c=6-14\\ -8c=-8\\ c=1\end{array}$

Putting the value of $c$ in equation -$\left(3\right)$, we get

$\begin{array}{c}3a-6c=6\\ 3a-6\left(1\right)=6\\ 3a=6+6\\ 3a=12\\ a=4\end{array}$

Step-3 – Substituting the value according to the coordinates in first equation.

$\begin{array}{c}2a+b-c=5\\ 2\left(4\right)+b-1=5\\ 8+b-1=5\\ b=5-7\\ b=-2\end{array}$