Solve by Addition/Elimination 3x-15y=9 , 2x-10y=6

$3 x - 15 y = 9$ , $2 x - 10 y = 6$

Step 1

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

$(- 2) \cdot (3 x - 15 y) = (- 2) (9)$

$(3) \cdot (2 x - 10 y) = (3) (6)$

Step 2

Simplify.

Simplify the left side.

Simplify $(- 2) \cdot (3 x - 15 y)$ .

Apply the distributive property.

$- 2 (3 x) - 2 (- 15 y) = (- 2) (9)$

$(3) \cdot (2 x - 10 y) = (3) (6)$

Multiply.

Multiply $3$ by $- 2$ .

$- 6 x - 2 (- 15 y) = (- 2) (9)$

$(3) \cdot (2 x - 10 y) = (3) (6)$

Multiply $- 15$ by $- 2$ .

$- 6 x + 30 y = (- 2) (9)$

$(3) \cdot (2 x - 10 y) = (3) (6)$

$- 6 x + 30 y = (- 2) (9)$

$(3) \cdot (2 x - 10 y) = (3) (6)$

$- 6 x + 30 y = (- 2) (9)$

$(3) \cdot (2 x - 10 y) = (3) (6)$

$- 6 x + 30 y = (- 2) (9)$

$(3) \cdot (2 x - 10 y) = (3) (6)$

Simplify the right side.

Multiply $- 2$ by $9$ .

$- 6 x + 30 y = - 18$

$(3) \cdot (2 x - 10 y) = (3) (6)$

$- 6 x + 30 y = - 18$

$(3) \cdot (2 x - 10 y) = (3) (6)$

Simplify the left side.

Simplify $(3) \cdot (2 x - 10 y)$ .

Apply the distributive property.

$- 6 x + 30 y = - 18$

$3 (2 x) + 3 (- 10 y) = (3) (6)$

Multiply.

Multiply $2$ by $3$ .

$- 6 x + 30 y = - 18$

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

Multiply $- 10$ by $3$ .

$- 6 x + 30 y = - 18$

$6 x - 30 y = (3) (6)$

$- 6 x + 30 y = - 18$

$6 x - 30 y = (3) (6)$

$- 6 x + 30 y = - 18$

$6 x - 30 y = (3) (6)$

$- 6 x + 30 y = - 18$

$6 x - 30 y = (3) (6)$

Simplify the right side.

Multiply $3$ by $6$ .

$- 6 x + 30 y = - 18$

$6 x - 30 y = 18$

$- 6 x + 30 y = - 18$

$6 x - 30 y = 18$

$- 6 x + 30 y = - 18$

$6 x - 30 y = 18$

Step 3

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

	$-$	$6$	$x$	$+$	$3$	$0$	$y$	$=$	$-$	$1$	$8$
$+$		$6$	$x$	$-$	$3$	$0$	$y$	$=$		$1$	$8$
							$0$	$=$			$0$

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 $y$ .

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

$30 y = - 18 + 6 x$

Divide each term in $30 y = - 18 + 6 x$ by $30$ and simplify.

Divide each term in $30 y = - 18 + 6 x$ by $30$ .

$\frac{30 y}{30} = \frac{- 18}{30} + \frac{6 x}{30}$

Simplify the left side.

Cancel the common factor of $30$ .

Cancel the common factor.

$\frac{30 y}{30} = \frac{- 18}{30} + \frac{6 x}{30}$

Divide $y$ by $1$ .

$y = \frac{- 18}{30} + \frac{6 x}{30}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 18$ and $30$ .

Factor $6$ out of $- 18$ .

$y = \frac{6 (- 3)}{30} + \frac{6 x}{30}$

Cancel the common factors.

Factor $6$ out of $30$ .

$y = \frac{6 \cdot - 3}{6 \cdot 5} + \frac{6 x}{30}$

Cancel the common factor.

$y = \frac{6 \cdot - 3}{6 \cdot 5} + \frac{6 x}{30}$

Rewrite the expression.

$y = \frac{- 3}{5} + \frac{6 x}{30}$

Move the negative in front of the fraction.

$y = - \frac{3}{5} + \frac{6 x}{30}$

Cancel the common factor of $6$ and $30$ .

Factor $6$ out of $6 x$ .

$y = - \frac{3}{5} + \frac{6 (x)}{30}$

Cancel the common factors.

Factor $6$ out of $30$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 6

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

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

Step 7

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

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

Step 8

Solve by Addition/Elimination 3x-15y=9 , 2x-10y=6

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