Solve by Addition/Elimination 2x+8y=6 , -5x-20y=-15

$2 x + 8 y = 6$ , $- 5 x - 20 y = - 15$

Step 1

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

$(5) \cdot (2 x + 8 y) = (5) (6)$

$(2) \cdot (- 5 x - 20 y) = (2) (- 15)$

Step 2

Simplify.

Simplify the left side.

Simplify $(5) \cdot (2 x + 8 y)$ .

Apply the distributive property.

$5 (2 x) + 5 (8 y) = (5) (6)$

$(2) \cdot (- 5 x - 20 y) = (2) (- 15)$

Multiply.

Multiply $2$ by $5$ .

$10 x + 5 (8 y) = (5) (6)$

$(2) \cdot (- 5 x - 20 y) = (2) (- 15)$

Multiply $8$ by $5$ .

$10 x + 40 y = (5) (6)$

$(2) \cdot (- 5 x - 20 y) = (2) (- 15)$

$10 x + 40 y = (5) (6)$

$(2) \cdot (- 5 x - 20 y) = (2) (- 15)$

$10 x + 40 y = (5) (6)$

$(2) \cdot (- 5 x - 20 y) = (2) (- 15)$

$10 x + 40 y = (5) (6)$

$(2) \cdot (- 5 x - 20 y) = (2) (- 15)$

Simplify the right side.

Multiply $5$ by $6$ .

$10 x + 40 y = 30$

$(2) \cdot (- 5 x - 20 y) = (2) (- 15)$

$10 x + 40 y = 30$

$(2) \cdot (- 5 x - 20 y) = (2) (- 15)$

Simplify the left side.

Simplify $(2) \cdot (- 5 x - 20 y)$ .

Apply the distributive property.

$10 x + 40 y = 30$

$2 (- 5 x) + 2 (- 20 y) = (2) (- 15)$

Multiply.

Multiply $- 5$ by $2$ .

$10 x + 40 y = 30$

$- 10 x + 2 (- 20 y) = (2) (- 15)$

Multiply $- 20$ by $2$ .

$10 x + 40 y = 30$

$- 10 x - 40 y = (2) (- 15)$

$10 x + 40 y = 30$

$- 10 x - 40 y = (2) (- 15)$

$10 x + 40 y = 30$

$- 10 x - 40 y = (2) (- 15)$

$10 x + 40 y = 30$

$- 10 x - 40 y = (2) (- 15)$

Simplify the right side.

Multiply $2$ by $- 15$ .

$10 x + 40 y = 30$

$- 10 x - 40 y = - 30$

$10 x + 40 y = 30$

$- 10 x - 40 y = - 30$

$10 x + 40 y = 30$

$- 10 x - 40 y = - 30$

Step 3

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

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

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 $y$ .

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

$40 y = 30 - 10 x$

Divide each term in $40 y = 30 - 10 x$ by $40$ and simplify.

Divide each term in $40 y = 30 - 10 x$ by $40$ .

$\frac{40 y}{40} = \frac{30}{40} + \frac{- 10 x}{40}$

Simplify the left side.

Cancel the common factor of $40$ .

Cancel the common factor.

$\frac{40 y}{40} = \frac{30}{40} + \frac{- 10 x}{40}$

Divide $y$ by $1$ .

$y = \frac{30}{40} + \frac{- 10 x}{40}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $30$ and $40$ .

Factor $10$ out of $30$ .

$y = \frac{10 (3)}{40} + \frac{- 10 x}{40}$

Cancel the common factors.

Factor $10$ out of $40$ .

$y = \frac{10 \cdot 3}{10 \cdot 4} + \frac{- 10 x}{40}$

Cancel the common factor.

$y = \frac{10 \cdot 3}{10 \cdot 4} + \frac{- 10 x}{40}$

Rewrite the expression.

$y = \frac{3}{4} + \frac{- 10 x}{40}$

Cancel the common factor of $- 10$ and $40$ .

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

$y = \frac{3}{4} + \frac{10 (- x)}{40}$

Cancel the common factors.

Factor $10$ out of $40$ .

$y = \frac{3}{4} + \frac{10 (- x)}{10 (4)}$

Cancel the common factor.

$y = \frac{3}{4} + \frac{10 (- x)}{10 \cdot 4}$

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Step 6

The solution is the set of ordered pairs that make $y = \frac{3}{4} - \frac{x}{4}$ true.

$(x, \frac{3}{4} - \frac{x}{4})$

Step 7

The result can be shown in multiple forms.

Point Form:

$(x, \frac{3}{4} - \frac{x}{4})$

Equation Form:

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

Step 8

Solve by Addition/Elimination 2x+8y=6 , -5x-20y=-15

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