Solve by Addition/Elimination 3x+y=5 2x-2y=-2

$3 x + y = 5$ $2 x - 2 y = - 2$

Step 1

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

$(2) \cdot (3 x + y) = (2) (5)$

$2 x - 2 y = - 2$

Step 2

Simplify.

Simplify the left side.

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

Apply the distributive property.

$2 (3 x) + 2 y = (2) (5)$

$2 x - 2 y = - 2$

Multiply $3$ by $2$ .

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

$2 x - 2 y = - 2$

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

$2 x - 2 y = - 2$

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

$2 x - 2 y = - 2$

Simplify the right side.

Multiply $2$ by $5$ .

$6 x + 2 y = 10$

$2 x - 2 y = - 2$

$6 x + 2 y = 10$

$2 x - 2 y = - 2$

$6 x + 2 y = 10$

$2 x - 2 y = - 2$

Step 3

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

	$6$	$x$	$+$	$2$	$y$	$=$	$1$	$0$
$+$	$2$	$x$	$-$	$2$	$y$	$=$	$-$	$2$
	$8$	$x$				$=$		$8$

Step 4

Divide each term in $8 x = 8$ by $8$ and simplify.

Divide each term in $8 x = 8$ by $8$ .

$\frac{8 x}{8} = \frac{8}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x}{8} = \frac{8}{8}$

Divide $x$ by $1$ .

$x = \frac{8}{8}$

Simplify the right side.

Divide $8$ by $8$ .

$x = 1$

Step 5

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

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

$6 (1) + 2 y = 10$

Multiply $6$ by $1$ .

$6 + 2 y = 10$

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

Subtract $6$ from both sides of the equation.

$2 y = 10 - 6$

Subtract $6$ from $10$ .

$2 y = 4$

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 6

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

$(1, 2)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(1, 2)$

Equation Form:

$x = 1, y = 2$

Step 8

Solve by Addition/Elimination 3x+y=5 2x-2y=-2

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