Solve by Addition/Elimination 5x-6y=4 -10x+12y=-8

$5 x - 6 y = 4$ $- 10 x + 12 y = - 8$

Step 1

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

$(2) \cdot (5 x - 6 y) = (2) (4)$

$- 10 x + 12 y = - 8$

Step 2

Simplify.

Simplify the left side.

Simplify $(2) \cdot (5 x - 6 y)$ .

Apply the distributive property.

$2 (5 x) + 2 (- 6 y) = (2) (4)$

$- 10 x + 12 y = - 8$

Multiply.

Multiply $5$ by $2$ .

$10 x + 2 (- 6 y) = (2) (4)$

$- 10 x + 12 y = - 8$

Multiply $- 6$ by $2$ .

$10 x - 12 y = (2) (4)$

$- 10 x + 12 y = - 8$

$10 x - 12 y = (2) (4)$

$- 10 x + 12 y = - 8$

$10 x - 12 y = (2) (4)$

$- 10 x + 12 y = - 8$

$10 x - 12 y = (2) (4)$

$- 10 x + 12 y = - 8$

Simplify the right side.

Multiply $2$ by $4$ .

$10 x - 12 y = 8$

$- 10 x + 12 y = - 8$

$10 x - 12 y = 8$

$- 10 x + 12 y = - 8$

$10 x - 12 y = 8$

$- 10 x + 12 y = - 8$

Step 3

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

		$1$	$0$	$x$	$-$	$1$	$2$	$y$	$=$		$8$
$+$	$-$	$1$	$0$	$x$	$+$	$1$	$2$	$y$	$=$	$-$	$8$
								$0$	$=$		$0$

Step 4

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

Infinite number of solutions

Step 5

Solve one of the equations for $y$ .

Subtract $10 x$ from both sides of the equation.

$- 12 y = 8 - 10 x$

Divide each term in $- 12 y = 8 - 10 x$ by $- 12$ and simplify.

Divide each term in $- 12 y = 8 - 10 x$ by $- 12$ .

$\frac{- 12 y}{- 12} = \frac{8}{- 12} + \frac{- 10 x}{- 12}$

Simplify the left side.

Cancel the common factor of $- 12$ .

Cancel the common factor.

$\frac{- 12 y}{- 12} = \frac{8}{- 12} + \frac{- 10 x}{- 12}$

Divide $y$ by $1$ .

$y = \frac{8}{- 12} + \frac{- 10 x}{- 12}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $8$ and $- 12$ .

Factor $4$ out of $8$ .

$y = \frac{4 (2)}{- 12} + \frac{- 10 x}{- 12}$

Cancel the common factors.

Factor $4$ out of $- 12$ .

$y = \frac{4 \cdot 2}{4 \cdot - 3} + \frac{- 10 x}{- 12}$

Cancel the common factor.

$y = \frac{4 \cdot 2}{4 \cdot - 3} + \frac{- 10 x}{- 12}$

Rewrite the expression.

$y = \frac{2}{- 3} + \frac{- 10 x}{- 12}$

Move the negative in front of the fraction.

$y = - \frac{2}{3} + \frac{- 10 x}{- 12}$

Cancel the common factor of $- 10$ and $- 12$ .

Factor $- 2$ out of $- 10 x$ .

$y = - \frac{2}{3} + \frac{- 2 (5 x)}{- 12}$

Cancel the common factors.

Factor $- 2$ out of $- 12$ .

$y = - \frac{2}{3} + \frac{- 2 (5 x)}{- 2 (6)}$

Cancel the common factor.

$y = - \frac{2}{3} + \frac{- 2 (5 x)}{- 2 \cdot 6}$

Rewrite the expression.

$y = - \frac{2}{3} + \frac{5 x}{6}$

Step 6

The solution is the set of ordered pairs that make $y = - \frac{2}{3} + \frac{5 x}{6}$ true.

$(x, - \frac{2}{3} + \frac{5 x}{6})$

Step 7

The result can be shown in multiple forms.

Point Form:

$(x, - \frac{2}{3} + \frac{5 x}{6})$

Equation Form:

$x = x, y = - \frac{2}{3} + \frac{5 x}{6}$

Step 8

Solve by Addition/Elimination 5x-6y=4 -10x+12y=-8

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