Solve by Addition/Elimination y=3x+8 12x-4y=-32

$y = 3 x + 8$ $12 x - 4 y = - 32$

Move all terms containing variables to the left side of the equation.

Subtract $3 x$ from both sides of the equation.

$y - 3 x = 8$

$12 x - 4 y = - 32$

Reorder $y$ and $- 3 x$ .

$- 3 x + y = 8$

$12 x - 4 y = - 32$

$- 3 x + y = 8$

$12 x - 4 y = - 32$

Multiply each equation by the value that makes the coefficients of $y$ opposite.

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

$12 x - 4 y = - 32$

Simplify.

Simplify the left side.

Simplify $(4) \cdot (- 3 x + y)$ .

Apply the distributive property.

$4 (- 3 x) + 4 y = (4) (8)$

$12 x - 4 y = - 32$

Multiply $- 3$ by $4$ .

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

$12 x - 4 y = - 32$

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

$12 x - 4 y = - 32$

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

$12 x - 4 y = - 32$

Simplify the right side.

Multiply $4$ by $8$ .

$- 12 x + 4 y = 32$

$12 x - 4 y = - 32$

$- 12 x + 4 y = 32$

$12 x - 4 y = - 32$

$- 12 x + 4 y = 32$

$12 x - 4 y = - 32$

Add the two equations together to eliminate $y$ from the system.

	$-$	$1$	$2$	$x$	$+$	$4$	$y$	$=$		$3$	$2$
$+$		$1$	$2$	$x$	$-$	$4$	$y$	$=$	$-$	$3$	$2$
							$0$	$=$			$0$

Since $0 = 0$ , the equations intersect at an infinite number of points.

Infinite number of solutions

Solve one of the equations for $y$ .

Add $12 x$ to both sides of the equation.

$4 y = 32 + 12 x$

Divide each term in $4 y = 32 + 12 x$ by $4$ and simplify.

Divide each term in $4 y = 32 + 12 x$ by $4$ .

$\frac{4 y}{4} = \frac{32}{4} + \frac{12 x}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{32}{4} + \frac{12 x}{4}$

Divide $y$ by $1$ .

$y = \frac{32}{4} + \frac{12 x}{4}$

Simplify the right side.

Simplify each term.

Divide $32$ by $4$ .

$y = 8 + \frac{12 x}{4}$

Cancel the common factor of $12$ and $4$ .

Factor $4$ out of $12 x$ .

$y = 8 + \frac{4 (3 x)}{4}$

Cancel the common factors.

Factor $4$ out of $4$ .

$y = 8 + \frac{4 (3 x)}{4 (1)}$

Cancel the common factor.

$y = 8 + \frac{4 (3 x)}{4 \cdot 1}$

Rewrite the expression.

$y = 8 + \frac{3 x}{1}$

Divide $3 x$ by $1$ .

$y = 8 + 3 x$

The solution is the set of ordered pairs that make $y = 8 + 3 x$ true.

$(x, 8 + 3 x)$

The result can be shown in multiple forms.

Point Form:

$(x, 8 + 3 x)$

Equation Form:

$x = x, y = 8 + 3 x$

Solve by Addition/Elimination y=3x+8 12x-4y=-32

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