Solve by Addition/Elimination 3x-6y=15 , 2x-4y=10

$3 x - 6 y = 15$ , $2 x - 4 y = 10$

Step 1

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

$(- 2) \cdot (3 x - 6 y) = (- 2) (15)$

$(3) \cdot (2 x - 4 y) = (3) (10)$

Step 2

Simplify.

Simplify the left side.

Simplify $(- 2) \cdot (3 x - 6 y)$ .

Apply the distributive property.

$- 2 (3 x) - 2 (- 6 y) = (- 2) (15)$

$(3) \cdot (2 x - 4 y) = (3) (10)$

Multiply.

Multiply $3$ by $- 2$ .

$- 6 x - 2 (- 6 y) = (- 2) (15)$

$(3) \cdot (2 x - 4 y) = (3) (10)$

Multiply $- 6$ by $- 2$ .

$- 6 x + 12 y = (- 2) (15)$

$(3) \cdot (2 x - 4 y) = (3) (10)$

$- 6 x + 12 y = (- 2) (15)$

$(3) \cdot (2 x - 4 y) = (3) (10)$

$- 6 x + 12 y = (- 2) (15)$

$(3) \cdot (2 x - 4 y) = (3) (10)$

$- 6 x + 12 y = (- 2) (15)$

$(3) \cdot (2 x - 4 y) = (3) (10)$

Simplify the right side.

Multiply $- 2$ by $15$ .

$- 6 x + 12 y = - 30$

$(3) \cdot (2 x - 4 y) = (3) (10)$

$- 6 x + 12 y = - 30$

$(3) \cdot (2 x - 4 y) = (3) (10)$

Simplify the left side.

Simplify $(3) \cdot (2 x - 4 y)$ .

Apply the distributive property.

$- 6 x + 12 y = - 30$

$3 (2 x) + 3 (- 4 y) = (3) (10)$

Multiply.

Multiply $2$ by $3$ .

$- 6 x + 12 y = - 30$

$6 x + 3 (- 4 y) = (3) (10)$

Multiply $- 4$ by $3$ .

$- 6 x + 12 y = - 30$

$6 x - 12 y = (3) (10)$

$- 6 x + 12 y = - 30$

$6 x - 12 y = (3) (10)$

$- 6 x + 12 y = - 30$

$6 x - 12 y = (3) (10)$

$- 6 x + 12 y = - 30$

$6 x - 12 y = (3) (10)$

Simplify the right side.

Multiply $3$ by $10$ .

$- 6 x + 12 y = - 30$

$6 x - 12 y = 30$

$- 6 x + 12 y = - 30$

$6 x - 12 y = 30$

$- 6 x + 12 y = - 30$

$6 x - 12 y = 30$

Step 3

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

	$-$	$6$	$x$	$+$	$1$	$2$	$y$	$=$	$-$	$3$	$0$
$+$		$6$	$x$	$-$	$1$	$2$	$y$	$=$		$3$	$0$
							$0$	$=$			$0$

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 $y$ .

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

$12 y = - 30 + 6 x$

Divide each term in $12 y = - 30 + 6 x$ by $12$ and simplify.

Divide each term in $12 y = - 30 + 6 x$ by $12$ .

$\frac{12 y}{12} = \frac{- 30}{12} + \frac{6 x}{12}$

Simplify the left side.

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 y}{12} = \frac{- 30}{12} + \frac{6 x}{12}$

Divide $y$ by $1$ .

$y = \frac{- 30}{12} + \frac{6 x}{12}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 30$ and $12$ .

Factor $6$ out of $- 30$ .

$y = \frac{6 (- 5)}{12} + \frac{6 x}{12}$

Cancel the common factors.

Factor $6$ out of $12$ .

$y = \frac{6 \cdot - 5}{6 \cdot 2} + \frac{6 x}{12}$

Cancel the common factor.

$y = \frac{6 \cdot - 5}{6 \cdot 2} + \frac{6 x}{12}$

Rewrite the expression.

$y = \frac{- 5}{2} + \frac{6 x}{12}$

Move the negative in front of the fraction.

$y = - \frac{5}{2} + \frac{6 x}{12}$

Cancel the common factor of $6$ and $12$ .

Factor $6$ out of $6 x$ .

$y = - \frac{5}{2} + \frac{6 (x)}{12}$

Cancel the common factors.

Factor $6$ out of $12$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 6

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

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

Step 7

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

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

Step 8

Solve by Addition/Elimination 3x-6y=15 , 2x-4y=10

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