Solve the System of Equations -a+4b+2c=-8 3a+b-4c=9 b=-1

$- a + 4 b + 2 c = - 8$ $3 a + b - 4 c = 9$ $b = - 1$

Replace all occurrences of $b$ with $- 1$ in each equation.

Replace all occurrences of $b$ in $- a + 4 b + 2 c = - 8$ with $- 1$ .

$- a + 4 (- 1) + 2 c = - 8$

$b = - 1$

$3 a + b - 4 c = 9$

Simplify the left side.

Multiply $4$ by $- 1$ .

$- a - 4 + 2 c = - 8$

$b = - 1$

$3 a + b - 4 c = 9$

$- a - 4 + 2 c = - 8$

$b = - 1$

$3 a + b - 4 c = 9$

Replace all occurrences of $b$ in $3 a + b - 4 c = 9$ with $- 1$ .

$3 a - 1 - 4 c = 9$

$- a - 4 + 2 c = - 8$

$b = - 1$

Simplify the left side.

Remove parentheses.

$3 a - 1 - 4 c = 9$

$- a - 4 + 2 c = - 8$

$b = - 1$

$3 a - 1 - 4 c = 9$

$- a - 4 + 2 c = - 8$

$b = - 1$

$3 a - 1 - 4 c = 9$

$- a - 4 + 2 c = - 8$

$b = - 1$

Solve for $a$ in $3 a - 1 - 4 c = 9$ .

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

Add $1$ to both sides of the equation.

$3 a - 4 c = 9 + 1$

$- a - 4 + 2 c = - 8$

$b = - 1$

Add $4 c$ to both sides of the equation.

$3 a = 9 + 1 + 4 c$

$- a - 4 + 2 c = - 8$

$b = - 1$

Add $9$ and $1$ .

$3 a = 10 + 4 c$

$- a - 4 + 2 c = - 8$

$b = - 1$

$3 a = 10 + 4 c$

$- a - 4 + 2 c = - 8$

$b = - 1$

Divide each term in $3 a = 10 + 4 c$ by $3$ and simplify.

Divide each term in $3 a = 10 + 4 c$ by $3$ .

$\frac{3 a}{3} = \frac{10}{3} + \frac{4 c}{3}$

$- a - 4 + 2 c = - 8$

$b = - 1$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 a}{3} = \frac{10}{3} + \frac{4 c}{3}$

$- a - 4 + 2 c = - 8$

$b = - 1$

Divide $a$ by $1$ .

$a = \frac{10}{3} + \frac{4 c}{3}$

$- a - 4 + 2 c = - 8$

$b = - 1$

$a = \frac{10}{3} + \frac{4 c}{3}$

$- a - 4 + 2 c = - 8$

$b = - 1$

$a = \frac{10}{3} + \frac{4 c}{3}$

$- a - 4 + 2 c = - 8$

$b = - 1$

$a = \frac{10}{3} + \frac{4 c}{3}$

$- a - 4 + 2 c = - 8$

$b = - 1$

$a = \frac{10}{3} + \frac{4 c}{3}$

$- a - 4 + 2 c = - 8$

$b = - 1$

Replace all occurrences of $a$ with $\frac{10}{3} + \frac{4 c}{3}$ in each equation.

Replace all occurrences of $a$ in $- a - 4 + 2 c = - 8$ with $\frac{10}{3} + \frac{4 c}{3}$ .

$- (\frac{10}{3} + \frac{4 c}{3}) - 4 + 2 c = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Simplify the left side.

Simplify $- (\frac{10}{3} + \frac{4 c}{3}) - 4 + 2 c$ .

Apply the distributive property.

$- \frac{10}{3} - \frac{4 c}{3} - 4 + 2 c = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

To write $- 4$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- \frac{4 c}{3} - \frac{10}{3} - 4 \cdot \frac{3}{3} + 2 c = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Combine $- 4$ and $\frac{3}{3}$ .

$- \frac{4 c}{3} - \frac{10}{3} + \frac{- 4 \cdot 3}{3} + 2 c = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Combine the numerators over the common denominator.

$- \frac{4 c}{3} + \frac{- 10 - 4 \cdot 3}{3} + 2 c = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Simplify the numerator.

Multiply $- 4$ by $3$ .

$- \frac{4 c}{3} + \frac{- 10 - 12}{3} + 2 c = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Subtract $12$ from $- 10$ .

$- \frac{4 c}{3} + \frac{- 22}{3} + 2 c = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

$- \frac{4 c}{3} + \frac{- 22}{3} + 2 c = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Move the negative in front of the fraction.

$- \frac{4 c}{3} - \frac{22}{3} + 2 c = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

To write $2 c$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- \frac{4 c}{3} + 2 c \cdot \frac{3}{3} - \frac{22}{3} = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Combine $2 c$ and $\frac{3}{3}$ .

$- \frac{4 c}{3} + \frac{2 c \cdot 3}{3} - \frac{22}{3} = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Combine the numerators over the common denominator.

$\frac{- 4 c + 2 c \cdot 3}{3} - \frac{22}{3} = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Combine the numerators over the common denominator.

$\frac{- 4 c + 2 c \cdot 3 - 22}{3} = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Multiply $3$ by $2$ .

$\frac{- 4 c + 6 c - 22}{3} = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Add $- 4 c$ and $6 c$ .

$\frac{2 c - 22}{3} = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Factor $2$ out of $2 c - 22$ .

Factor $2$ out of $2 c$ .

$\frac{2 (c) - 22}{3} = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Factor $2$ out of $- 22$ .

$\frac{2 c + 2 \cdot - 11}{3} = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Factor $2$ out of $2 c + 2 \cdot - 11$ .

$\frac{2 (c - 11)}{3} = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

$\frac{2 (c - 11)}{3} = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

$\frac{2 (c - 11)}{3} = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

$\frac{2 (c - 11)}{3} = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

$\frac{2 (c - 11)}{3} = - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Solve for $c$ in $\frac{2 (c - 11)}{3} = - 8$ .

Multiply both sides of the equation by $\frac{3}{2}$ .

$\frac{3}{2} \cdot \frac{2 (c - 11)}{3} = \frac{3}{2} \cdot - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{3}{2} \cdot \frac{2 (c - 11)}{3}$ .

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3}{2} \cdot \frac{2 (c - 11)}{3} = \frac{3}{2} \cdot - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Rewrite the expression.

$\frac{1}{2} \cdot (2 (c - 11)) = \frac{3}{2} \cdot - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

$\frac{1}{2} \cdot (2 (c - 11)) = \frac{3}{2} \cdot - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{1}{2} \cdot (2 (c - 11)) = \frac{3}{2} \cdot - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Rewrite the expression.

$c - 11 = \frac{3}{2} \cdot - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

$c - 11 = \frac{3}{2} \cdot - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

$c - 11 = \frac{3}{2} \cdot - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

$c - 11 = \frac{3}{2} \cdot - 8$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Simplify the right side.

Simplify $\frac{3}{2} \cdot - 8$ .

Cancel the common factor of $2$ .

Factor $2$ out of $- 8$ .

$c - 11 = \frac{3}{2} \cdot (2 (- 4))$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Cancel the common factor.

$c - 11 = \frac{3}{2} \cdot (2 \cdot - 4)$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Rewrite the expression.

$c - 11 = 3 \cdot - 4$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

$c - 11 = 3 \cdot - 4$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Multiply $3$ by $- 4$ .

$c - 11 = - 12$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

$c - 11 = - 12$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

$c - 11 = - 12$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

$c - 11 = - 12$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

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

Add $11$ to both sides of the equation.

$c = - 12 + 11$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Add $- 12$ and $11$ .

$c = - 1$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

$c = - 1$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

$c = - 1$

$a = \frac{10}{3} + \frac{4 c}{3}$

$b = - 1$

Replace all occurrences of $c$ with $- 1$ in each equation.

Replace all occurrences of $c$ in $a = \frac{10}{3} + \frac{4 c}{3}$ with $- 1$ .

$a = \frac{10}{3} + \frac{4 (- 1)}{3}$

$c = - 1$

$b = - 1$

Simplify the right side.

Simplify $\frac{10}{3} + \frac{4 (- 1)}{3}$ .

Combine the numerators over the common denominator.

$a = \frac{10 + 4 (- 1)}{3}$

$c = - 1$

$b = - 1$

Simplify the expression.

Multiply $4$ by $- 1$ .

$a = \frac{10 - 4}{3}$

$c = - 1$

$b = - 1$

Subtract $4$ from $10$ .

$a = \frac{6}{3}$

$c = - 1$

$b = - 1$

Divide $6$ by $3$ .

$a = 2$

$c = - 1$

$b = - 1$

$a = 2$

$c = - 1$

$b = - 1$

$a = 2$

$c = - 1$

$b = - 1$

$a = 2$

$c = - 1$

$b = - 1$

$a = 2$

$c = - 1$

$b = - 1$

List all of the solutions.

$a = 2, c = - 1, b = - 1$

Solve the System of Equations -a+4b+2c=-8 3a+b-4c=9 b=-1

Related Questions

Support LCMGCF.com by disabling your adblocker.