Solve the System of Equations 2x+2y=-8 and 3x+3y=-12

$2 x + 2 y = - 8$ and $3 x + 3 y = - 12$

Solve for $x$ in $2 x + 2 y = - 8$ .

Subtract $2 y$ from both sides of the equation.

$2 x = - 8 - 2 y$

$3 x + 3 y = - 12$

Divide each term in $2 x = - 8 - 2 y$ by $2$ and simplify.

Divide each term in $2 x = - 8 - 2 y$ by $2$ .

$\frac{2 x}{2} = \frac{- 8}{2} + \frac{- 2 y}{2}$

$3 x + 3 y = - 12$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 8}{2} + \frac{- 2 y}{2}$

$3 x + 3 y = - 12$

Divide $x$ by $1$ .

$x = \frac{- 8}{2} + \frac{- 2 y}{2}$

$3 x + 3 y = - 12$

$x = \frac{- 8}{2} + \frac{- 2 y}{2}$

$3 x + 3 y = - 12$

$x = \frac{- 8}{2} + \frac{- 2 y}{2}$

$3 x + 3 y = - 12$

Simplify the right side.

Simplify each term.

Divide $- 8$ by $2$ .

$x = - 4 + \frac{- 2 y}{2}$

$3 x + 3 y = - 12$

Cancel the common factor of $- 2$ and $2$ .

Factor $2$ out of $- 2 y$ .

$x = - 4 + \frac{2 (- y)}{2}$

$3 x + 3 y = - 12$

Cancel the common factors.

Factor $2$ out of $2$ .

$x = - 4 + \frac{2 (- y)}{2 (1)}$

$3 x + 3 y = - 12$

Cancel the common factor.

$x = - 4 + \frac{2 (- y)}{2 \cdot 1}$

$3 x + 3 y = - 12$

Rewrite the expression.

$x = - 4 + \frac{- y}{1}$

$3 x + 3 y = - 12$

Divide $- y$ by $1$ .

$x = - 4 - y$

$3 x + 3 y = - 12$

$x = - 4 - y$

$3 x + 3 y = - 12$

$x = - 4 - y$

$3 x + 3 y = - 12$

$x = - 4 - y$

$3 x + 3 y = - 12$

$x = - 4 - y$

$3 x + 3 y = - 12$

$x = - 4 - y$

$3 x + 3 y = - 12$

$x = - 4 - y$

$3 x + 3 y = - 12$

Replace all occurrences of $x$ with $- 4 - y$ in each equation.

Replace all occurrences of $x$ in $3 x + 3 y = - 12$ with $- 4 - y$ .

$3 (- 4 - y) + 3 y = - 12$

$x = - 4 - y$

Simplify the left side.

Simplify $3 (- 4 - y) + 3 y$ .

Simplify each term.

Apply the distributive property.

$3 \cdot - 4 + 3 (- y) + 3 y = - 12$

$x = - 4 - y$

Multiply $3$ by $- 4$ .

$- 12 + 3 (- y) + 3 y = - 12$

$x = - 4 - y$

Multiply $- 1$ by $3$ .

$- 12 - 3 y + 3 y = - 12$

$x = - 4 - y$

$- 12 - 3 y + 3 y = - 12$

$x = - 4 - y$

Combine the opposite terms in $- 12 - 3 y + 3 y$ .

Add $- 3 y$ and $3 y$ .

$- 12 + 0 = - 12$

$x = - 4 - y$

Add $- 12$ and $0$ .

$- 12 = - 12$

$x = - 4 - y$

$- 12 = - 12$

$x = - 4 - y$

$- 12 = - 12$

$x = - 4 - y$

$- 12 = - 12$

$x = - 4 - y$

$- 12 = - 12$

$x = - 4 - y$

Remove any equations from the system that are always true.

$x = - 4 - y$

Solve the System of Equations 2x+2y=-8 and 3x+3y=-12

Related Questions

Support LCMGCF.com by disabling your adblocker.