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Expert-verified Found in: Page 228 ### Algebra 2

Book edition Middle English Edition
Author(s) Carter
Pages 804 pages
ISBN 9780079039903 # Solve the following system of equations?$\begin{array}{c}a+b+c=6\\ 2a-b+3c=16\\ a+3b-2c=-6\end{array}$

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

See the step by step solution

## Step 1. Write down the given information.

The given system of equations are,

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

## Step 2. Calculation.

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

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

Subtract (3) from (1) gives,

$\begin{array}{c}\left(a+b+c\right)-\left(a+3b-2c\right)=6+6\\ -2b+3c=12\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}....\left(5\right)\end{array}$

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

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

Plugging $b=0$ in (4) gives $c=4$.

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

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

## Step 3. Conclusion.

The solution for the system of equations is $a=2,b=0,c=4$ . ### Want to see more solutions like these? 