Solve the System of Equations x+2y=-4 4y=3x+12

$x + 2 y = - 4$ $4 y = 3 x + 12$

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

$x = - 4 - 2 y$

$4 y = 3 x + 12$

Replace all occurrences of $x$ with $- 4 - 2 y$ in each equation.

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

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

$x = - 4 - 2 y$

Simplify the right side.

Simplify $3 (- 4 - 2 y) + 12$ .

Simplify each term.

Apply the distributive property.

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

$x = - 4 - 2 y$

Multiply $3$ by $- 4$ .

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

$x = - 4 - 2 y$

Multiply $- 2$ by $3$ .

$4 y = - 12 - 6 y + 12$

$x = - 4 - 2 y$

$4 y = - 12 - 6 y + 12$

$x = - 4 - 2 y$

Combine the opposite terms in $- 12 - 6 y + 12$ .

Add $- 12$ and $12$ .

$4 y = - 6 y + 0$

$x = - 4 - 2 y$

Add $- 6 y$ and $0$ .

$4 y = - 6 y$

$x = - 4 - 2 y$

$4 y = - 6 y$

$x = - 4 - 2 y$

$4 y = - 6 y$

$x = - 4 - 2 y$

$4 y = - 6 y$

$x = - 4 - 2 y$

$4 y = - 6 y$

$x = - 4 - 2 y$

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

Move all terms containing $y$ to the left side of the equation.

Add $6 y$ to both sides of the equation.

$4 y + 6 y = 0$

$x = - 4 - 2 y$

Add $4 y$ and $6 y$ .

$10 y = 0$

$x = - 4 - 2 y$

$10 y = 0$

$x = - 4 - 2 y$

Divide each term in $10 y = 0$ by $10$ and simplify.

Divide each term in $10 y = 0$ by $10$ .

$\frac{10 y}{10} = \frac{0}{10}$

$x = - 4 - 2 y$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 y}{10} = \frac{0}{10}$

$x = - 4 - 2 y$

Divide $y$ by $1$ .

$y = \frac{0}{10}$

$x = - 4 - 2 y$

$y = \frac{0}{10}$

$x = - 4 - 2 y$

$y = \frac{0}{10}$

$x = - 4 - 2 y$

Simplify the right side.

Divide $0$ by $10$ .

$y = 0$

$x = - 4 - 2 y$

$y = 0$

$x = - 4 - 2 y$

$y = 0$

$x = - 4 - 2 y$

$y = 0$

$x = - 4 - 2 y$

Replace all occurrences of $y$ with $0$ in each equation.

Replace all occurrences of $y$ in $x = - 4 - 2 y$ with $0$ .

$x = - 4 - 2 \cdot 0$

$y = 0$

Simplify the right side.

Simplify $- 4 - 2 \cdot 0$ .

Multiply $- 2$ by $0$ .

$x = - 4 + 0$

$y = 0$

Add $- 4$ and $0$ .

$x = - 4$

$y = 0$

$x = - 4$

$y = 0$

$x = - 4$

$y = 0$

$x = - 4$

$y = 0$

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(- 4, 0)$

The result can be shown in multiple forms.

Point Form:

$(- 4, 0)$

Equation Form:

$x = - 4, y = 0$

Solve the System of Equations x+2y=-4 4y=3x+12

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