Q73. - Algebra 2

Question

Solve the following system of equations?

$\begin{array}{c}a+b+c=6\\ 2a-b+3c=16\\ a+3b-2c=-6\end{array}$

Step-by-Step Solution

Answer

The solution for the system of equations is $a=2,b=0,c=4$ .

Step 1. Write down the given information.

The given system of equations are,

$\begin{array}{c}a+b+c=\mathrm{6....}\left(1\right)\\ 2a-b+3c=\mathrm{16....}\left(2\right)\\ a+3b-2c=-\mathrm{6....}\left(3\right)\end{array}$

Step 2. Calculation.

Multiply (1) by 2 and subtract from (2) gives,

$\begin{array}{c}(2a-b+3c)-(2a+2b+2c)=16-12\\ (2a-b+3c)-(2a+2b+2c)=4\\ -3b+c=\mathrm{4....}\left(4\right)\end{array}$

Subtract (3) from (1) gives,

$\begin{array}{c}(a+b+c)-(a+3b-2c)=6+6\\ -2b+3c=12\mathrm{....}\left(5\right)\end{array}$

Multiply (4) by 3 and subtract from (5) gives,

$\begin{array}{c}(-2b+3c)-(-9b+3c)=12-12\\ 7b=0\\ b=0\end{array}$

Plugging $b=0$  in (4) gives  $c=4$.

Again plugging $b=0$  and $c=4$ in (1),

$\begin{array}{c}a+b+c=\mathrm{6....}\left(\text{From }\left(1\right)\right)\\ a+\left(0\right)+\left(4\right)=\mathrm{6....}\left(\begin{array}{l}\text{Plugging}\\ b=0,c=4\end{array}\right)\\ a=2\end{array}$

Step 3. Conclusion.

The solution for the system of equations is $a=2,b=0,c=4$ .