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

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

Step 1

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

$x + y = 1$

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

Step 2

Simplify.

Simplify the left side.

Simplify $(- 1) \cdot (x - y)$ .

Apply the distributive property.

$x + y = 1$

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

Rewrite $- 1 x$ as $- x$ .

$x + y = 1$

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

Multiply $- 1 (- y)$ .

Multiply $- 1$ by $- 1$ .

$x + y = 1$

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

Multiply $y$ by $1$ .

$x + y = 1$

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

$x + y = 1$

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

$x + y = 1$

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

$x + y = 1$

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

Simplify the right side.

Multiply $- 1$ by $3$ .

$x + y = 1$

$- x + y = - 3$

$x + y = 1$

$- x + y = - 3$

$x + y = 1$

$- x + y = - 3$

Step 3

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

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

Step 4

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Divide $- 2$ by $2$ .

$y = - 1$

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 $x$ .

$x - 1 = 1$

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

Add $1$ to both sides of the equation.

$x = 1 + 1$

Add $1$ and $1$ .

$x = 2$

Step 6

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

$(2, - 1)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(2, - 1)$

Equation Form:

$x = 2, y = - 1$

Step 8

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

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