Solve the System of Equations a-2b+c=8 2a+b-c=0 3a-6b+3c=24

$a - 2 b + c = 8$ $2 a + b - c = 0$ $3 a - 6 b + 3 c = 24$

Step 1

Move all terms not containing $a$ to the right side of the equation.

Add $2 b$ to both sides of the equation.

$a + c = 8 + 2 b$

$2 a + b - c = 0$

$3 a - 6 b + 3 c = 24$

Subtract $c$ from both sides of the equation.

$a = 8 + 2 b - c$

$2 a + b - c = 0$

$3 a - 6 b + 3 c = 24$

$a = 8 + 2 b - c$

$2 a + b - c = 0$

$3 a - 6 b + 3 c = 24$

Step 2

Replace all occurrences of $a$ with $8 + 2 b - c$ in each equation.

Replace all occurrences of $a$ in $2 a + b - c = 0$ with $8 + 2 b - c$ .

$2 (8 + 2 b - c) + b - c = 0$

$a = 8 + 2 b - c$

$3 a - 6 b + 3 c = 24$

Simplify the left side.

Simplify $2 (8 + 2 b - c) + b - c$ .

Simplify each term.

Apply the distributive property.

$2 \cdot 8 + 2 (2 b) + 2 (- c) + b - c = 0$

$a = 8 + 2 b - c$

$3 a - 6 b + 3 c = 24$

Simplify.

Multiply $2$ by $8$ .

$16 + 2 (2 b) + 2 (- c) + b - c = 0$

$a = 8 + 2 b - c$

$3 a - 6 b + 3 c = 24$

Multiply $2$ by $2$ .

$16 + 4 b + 2 (- c) + b - c = 0$

$a = 8 + 2 b - c$

$3 a - 6 b + 3 c = 24$

Multiply $- 1$ by $2$ .

$16 + 4 b - 2 c + b - c = 0$

$a = 8 + 2 b - c$

$3 a - 6 b + 3 c = 24$

$16 + 4 b - 2 c + b - c = 0$

$a = 8 + 2 b - c$

$3 a - 6 b + 3 c = 24$

$16 + 4 b - 2 c + b - c = 0$

$a = 8 + 2 b - c$

$3 a - 6 b + 3 c = 24$

Simplify by adding terms.

Add $4 b$ and $b$ .

$16 + 5 b - 2 c - c = 0$

$a = 8 + 2 b - c$

$3 a - 6 b + 3 c = 24$

Subtract $c$ from $- 2 c$ .

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

$3 a - 6 b + 3 c = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

$3 a - 6 b + 3 c = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

$3 a - 6 b + 3 c = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

$3 a - 6 b + 3 c = 24$

Replace all occurrences of $a$ in $3 a - 6 b + 3 c = 24$ with $8 + 2 b - c$ .

$3 (8 + 2 b - c) - 6 b + 3 c = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

Simplify the left side.

Simplify $3 (8 + 2 b - c) - 6 b + 3 c$ .

Simplify each term.

Apply the distributive property.

$3 \cdot 8 + 3 (2 b) + 3 (- c) - 6 b + 3 c = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

Simplify.

Multiply $3$ by $8$ .

$24 + 3 (2 b) + 3 (- c) - 6 b + 3 c = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

Multiply $2$ by $3$ .

$24 + 6 b + 3 (- c) - 6 b + 3 c = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

Multiply $- 1$ by $3$ .

$24 + 6 b - 3 c - 6 b + 3 c = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

$24 + 6 b - 3 c - 6 b + 3 c = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

$24 + 6 b - 3 c - 6 b + 3 c = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

Combine the opposite terms in $24 + 6 b - 3 c - 6 b + 3 c$ .

Subtract $6 b$ from $6 b$ .

$24 - 3 c + 0 + 3 c = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

Add $24 - 3 c$ and $0$ .

$24 - 3 c + 3 c = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

Add $- 3 c$ and $3 c$ .

$24 + 0 = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

Add $24$ and $0$ .

$24 = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

$24 = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

$24 = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

$24 = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

$24 = 24$

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

Step 3

Remove any equations from the system that are always true.

$16 + 5 b - 3 c = 0$

$a = 8 + 2 b - c$

Solve the System of Equations a-2b+c=8 2a+b-c=0 3a-6b+3c=24

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