Solve the System of Equations 2x-4y=6 -4x+8y=-12

$2 x - 4 y = 6$ $- 4 x + 8 y = - 12$

Solve for $x$ in $2 x - 4 y = 6$ .

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

$2 x = 6 + 4 y$

$- 4 x + 8 y = - 12$

Divide each term in $2 x = 6 + 4 y$ by $2$ and simplify.

Divide each term in $2 x = 6 + 4 y$ by $2$ .

$\frac{2 x}{2} = \frac{6}{2} + \frac{4 y}{2}$

$- 4 x + 8 y = - 12$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{6}{2} + \frac{4 y}{2}$

$- 4 x + 8 y = - 12$

Divide $x$ by $1$ .

$x = \frac{6}{2} + \frac{4 y}{2}$

$- 4 x + 8 y = - 12$

$x = \frac{6}{2} + \frac{4 y}{2}$

$- 4 x + 8 y = - 12$

$x = \frac{6}{2} + \frac{4 y}{2}$

$- 4 x + 8 y = - 12$

Simplify the right side.

Simplify each term.

Divide $6$ by $2$ .

$x = 3 + \frac{4 y}{2}$

$- 4 x + 8 y = - 12$

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4 y$ .

$x = 3 + \frac{2 (2 y)}{2}$

$- 4 x + 8 y = - 12$

Cancel the common factors.

Factor $2$ out of $2$ .

$x = 3 + \frac{2 (2 y)}{2 (1)}$

$- 4 x + 8 y = - 12$

Cancel the common factor.

$x = 3 + \frac{2 (2 y)}{2 \cdot 1}$

$- 4 x + 8 y = - 12$

Rewrite the expression.

$x = 3 + \frac{2 y}{1}$

$- 4 x + 8 y = - 12$

Divide $2 y$ by $1$ .

$x = 3 + 2 y$

$- 4 x + 8 y = - 12$

$x = 3 + 2 y$

$- 4 x + 8 y = - 12$

$x = 3 + 2 y$

$- 4 x + 8 y = - 12$

$x = 3 + 2 y$

$- 4 x + 8 y = - 12$

$x = 3 + 2 y$

$- 4 x + 8 y = - 12$

$x = 3 + 2 y$

$- 4 x + 8 y = - 12$

$x = 3 + 2 y$

$- 4 x + 8 y = - 12$

Replace all occurrences of $x$ with $3 + 2 y$ in each equation.

Replace all occurrences of $x$ in $- 4 x + 8 y = - 12$ with $3 + 2 y$ .

$- 4 (3 + 2 y) + 8 y = - 12$

$x = 3 + 2 y$

Simplify the left side.

Simplify $- 4 (3 + 2 y) + 8 y$ .

Simplify each term.

Apply the distributive property.

$- 4 \cdot 3 - 4 (2 y) + 8 y = - 12$

$x = 3 + 2 y$

Multiply $- 4$ by $3$ .

$- 12 - 4 (2 y) + 8 y = - 12$

$x = 3 + 2 y$

Multiply $2$ by $- 4$ .

$- 12 - 8 y + 8 y = - 12$

$x = 3 + 2 y$

$- 12 - 8 y + 8 y = - 12$

$x = 3 + 2 y$

Combine the opposite terms in $- 12 - 8 y + 8 y$ .

Add $- 8 y$ and $8 y$ .

$- 12 + 0 = - 12$

$x = 3 + 2 y$

Add $- 12$ and $0$ .

$- 12 = - 12$

$x = 3 + 2 y$

$- 12 = - 12$

$x = 3 + 2 y$

$- 12 = - 12$

$x = 3 + 2 y$

$- 12 = - 12$

$x = 3 + 2 y$

$- 12 = - 12$

$x = 3 + 2 y$

Remove any equations from the system that are always true.

$x = 3 + 2 y$

Solve the System of Equations 2x-4y=6 -4x+8y=-12

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