Solve by Addition/Elimination x+2y=-1 5x-4y=16

$x + 2 y = - 1$ $5 x - 4 y = 16$

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

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

$5 x - 4 y = 16$

Simplify.

Simplify the left side.

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

Apply the distributive property.

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

$5 x - 4 y = 16$

Multiply $2$ by $2$ .

$2 x + 4 y = (2) (- 1)$

$5 x - 4 y = 16$

$2 x + 4 y = (2) (- 1)$

$5 x - 4 y = 16$

$2 x + 4 y = (2) (- 1)$

$5 x - 4 y = 16$

Simplify the right side.

Multiply $2$ by $- 1$ .

$2 x + 4 y = - 2$

$5 x - 4 y = 16$

$2 x + 4 y = - 2$

$5 x - 4 y = 16$

$2 x + 4 y = - 2$

$5 x - 4 y = 16$

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

	$2$	$x$	$+$	$4$	$y$	$=$	$-$	$2$
$+$	$5$	$x$	$-$	$4$	$y$	$=$	$1$	$6$
	$7$	$x$				$=$	$1$	$4$

Divide each term in $7 x = 14$ by $7$ and simplify.

Divide each term in $7 x = 14$ by $7$ .

$\frac{7 x}{7} = \frac{14}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{14}{7}$

Divide $x$ by $1$ .

$x = \frac{14}{7}$

Simplify the right side.

Divide $14$ by $7$ .

$x = 2$

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 $y$ .

$2 (2) + 4 y = - 2$

Multiply $2$ by $2$ .

$4 + 4 y = - 2$

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

Subtract $4$ from both sides of the equation.

$4 y = - 2 - 4$

Subtract $4$ from $- 2$ .

$4 y = - 6$

Divide each term in $4 y = - 6$ by $4$ and simplify.

Divide each term in $4 y = - 6$ by $4$ .

$\frac{4 y}{4} = \frac{- 6}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{- 6}{4}$

Divide $y$ by $1$ .

$y = \frac{- 6}{4}$

Simplify the right side.

Cancel the common factor of $- 6$ and $4$ .

Factor $2$ out of $- 6$ .

$y = \frac{2 (- 3)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

$y = \frac{2 \cdot - 3}{2 \cdot 2}$

Cancel the common factor.

$y = \frac{2 \cdot - 3}{2 \cdot 2}$

Rewrite the expression.

$y = \frac{- 3}{2}$

Move the negative in front of the fraction.

$y = - \frac{3}{2}$

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

$(2, - \frac{3}{2})$

The result can be shown in multiple forms.

Point Form:

$(2, - \frac{3}{2})$

Equation Form:

$x = 2, y = - \frac{3}{2}$

Solve by Addition/Elimination x+2y=-1 5x-4y=16

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