Solve the System of Equations 2x+4y=12 y=1/4x-3

$2 x + 4 y = 12$ $y = \frac{1}{4} x - 3$

Combine $\frac{1}{4}$ and $x$ .

$y = \frac{x}{4} - 3$

$2 x + 4 y = 12$

Replace all occurrences of $y$ with $\frac{x}{4} - 3$ in each equation.

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

$2 x + 4 (\frac{x}{4} - 3) = 12$

$y = \frac{x}{4} - 3$

Simplify the left side.

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

Simplify each term.

Apply the distributive property.

$2 x + 4 (\frac{x}{4}) + 4 \cdot - 3 = 12$

$y = \frac{x}{4} - 3$

Cancel the common factor of $4$ .

Cancel the common factor.

$2 x + 4 (\frac{x}{4}) + 4 \cdot - 3 = 12$

$y = \frac{x}{4} - 3$

Rewrite the expression.

$2 x + x + 4 \cdot - 3 = 12$

$y = \frac{x}{4} - 3$

$2 x + x + 4 \cdot - 3 = 12$

$y = \frac{x}{4} - 3$

Multiply $4$ by $- 3$ .

$2 x + x - 12 = 12$

$y = \frac{x}{4} - 3$

$2 x + x - 12 = 12$

$y = \frac{x}{4} - 3$

Add $2 x$ and $x$ .

$3 x - 12 = 12$

$y = \frac{x}{4} - 3$

$3 x - 12 = 12$

$y = \frac{x}{4} - 3$

$3 x - 12 = 12$

$y = \frac{x}{4} - 3$

$3 x - 12 = 12$

$y = \frac{x}{4} - 3$

Solve for $x$ in $3 x - 12 = 12$ .

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

Add $12$ to both sides of the equation.

$3 x = 12 + 12$

$y = \frac{x}{4} - 3$

Add $12$ and $12$ .

$3 x = 24$

$y = \frac{x}{4} - 3$

$3 x = 24$

$y = \frac{x}{4} - 3$

Divide each term in $3 x = 24$ by $3$ and simplify.

Divide each term in $3 x = 24$ by $3$ .

$\frac{3 x}{3} = \frac{24}{3}$

$y = \frac{x}{4} - 3$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{24}{3}$

$y = \frac{x}{4} - 3$

Divide $x$ by $1$ .

$x = \frac{24}{3}$

$y = \frac{x}{4} - 3$

$x = \frac{24}{3}$

$y = \frac{x}{4} - 3$

$x = \frac{24}{3}$

$y = \frac{x}{4} - 3$

Simplify the right side.

Divide $24$ by $3$ .

$x = 8$

$y = \frac{x}{4} - 3$

$x = 8$

$y = \frac{x}{4} - 3$

$x = 8$

$y = \frac{x}{4} - 3$

$x = 8$

$y = \frac{x}{4} - 3$

Replace all occurrences of $x$ with $8$ in each equation.

Replace all occurrences of $x$ in $y = \frac{x}{4} - 3$ with $8$ .

$y = \frac{8}{4} - 3$

$x = 8$

Simplify the right side.

Simplify $\frac{8}{4} - 3$ .

Divide $8$ by $4$ .

$y = 2 - 3$

$x = 8$

Subtract $3$ from $2$ .

$y = - 1$

$x = 8$

$y = - 1$

$x = 8$

$y = - 1$

$x = 8$

$y = - 1$

$x = 8$

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

$(8, - 1)$

The result can be shown in multiple forms.

Point Form:

$(8, - 1)$

Equation Form:

$x = 8, y = - 1$

Solve the System of Equations 2x+4y=12 y=1/4x-3

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