Solve by Addition/Elimination x+3y=1 , 2x+4y=-2

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

Step 1

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

$(- 2) \cdot (x + 3 y) = (- 2) (1)$

$2 x + 4 y = - 2$

Step 2

Simplify.

Simplify the left side.

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

Apply the distributive property.

$- 2 x - 2 (3 y) = (- 2) (1)$

$2 x + 4 y = - 2$

Multiply $3$ by $- 2$ .

$- 2 x - 6 y = (- 2) (1)$

$2 x + 4 y = - 2$

$- 2 x - 6 y = (- 2) (1)$

$2 x + 4 y = - 2$

$- 2 x - 6 y = (- 2) (1)$

$2 x + 4 y = - 2$

Simplify the right side.

Multiply $- 2$ by $1$ .

$- 2 x - 6 y = - 2$

$2 x + 4 y = - 2$

$- 2 x - 6 y = - 2$

$2 x + 4 y = - 2$

$- 2 x - 6 y = - 2$

$2 x + 4 y = - 2$

Step 3

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

	$-$	$2$	$x$	$-$	$6$	$y$	$=$	$-$	$2$
$+$		$2$	$x$	$+$	$4$	$y$	$=$	$-$	$2$
				$-$	$2$	$y$	$=$	$-$	$4$

Step 4

Divide each term in $- 2 y = - 4$ by $- 2$ and simplify.

Divide each term in $- 2 y = - 4$ by $- 2$ .

$\frac{- 2 y}{- 2} = \frac{- 4}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 y}{- 2} = \frac{- 4}{- 2}$

Divide $y$ by $1$ .

$y = \frac{- 4}{- 2}$

Simplify the right side.

Divide $- 4$ by $- 2$ .

$y = 2$

Step 5

Substitute the value found for $y$ into one of the original equations, then solve for $x$ .

Substitute the value found for $y$ into one of the original equations to solve for $x$ .

$- 2 x - 6 \cdot 2 = - 2$

Multiply $- 6$ by $2$ .

$- 2 x - 12 = - 2$

Move all terms not containing $x$ to the right side of the equation.

Add $12$ to both sides of the equation.

$- 2 x = - 2 + 12$

Add $- 2$ and $12$ .

$- 2 x = 10$

Divide each term in $- 2 x = 10$ by $- 2$ and simplify.

Divide each term in $- 2 x = 10$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{10}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{10}{- 2}$

Divide $x$ by $1$ .

$x = \frac{10}{- 2}$

Simplify the right side.

Divide $10$ by $- 2$ .

$x = - 5$

Step 6

The solution to the independent system of equations can be represented as a point.

$(- 5, 2)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(- 5, 2)$

Equation Form:

$x = - 5, y = 2$

Step 8

Solve by Addition/Elimination x+3y=1 , 2x+4y=-2

Related Questions

Support LCMGCF.com by disabling your adblocker.