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

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

Subtract $2 y$ from both sides of the equation.

$x - 2 y = - 3$

$2 x - 3 y = - 5$

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

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

$2 x - 3 y = - 5$

Simplify.

Simplify the left side.

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

Apply the distributive property.

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

$2 x - 3 y = - 5$

Multiply $- 2$ by $- 2$ .

$- 2 x + 4 y = (- 2) (- 3)$

$2 x - 3 y = - 5$

$- 2 x + 4 y = (- 2) (- 3)$

$2 x - 3 y = - 5$

$- 2 x + 4 y = (- 2) (- 3)$

$2 x - 3 y = - 5$

Simplify the right side.

Multiply $- 2$ by $- 3$ .

$- 2 x + 4 y = 6$

$2 x - 3 y = - 5$

$- 2 x + 4 y = 6$

$2 x - 3 y = - 5$

$- 2 x + 4 y = 6$

$2 x - 3 y = - 5$

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

	$-$	$2$	$x$	$+$	$4$	$y$	$=$		$6$
$+$		$2$	$x$	$-$	$3$	$y$	$=$	$-$	$5$
						$y$	$=$		$1$

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 $x$ .

$- 2 x + 4 (1) = 6$

Multiply $4$ by $1$ .

$- 2 x + 4 = 6$

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

Subtract $4$ from both sides of the equation.

$- 2 x = 6 - 4$

Subtract $4$ from $6$ .

$- 2 x = 2$

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

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

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

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $2$ by $- 2$ .

$x = - 1$

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

$(- 1, 1)$

The result can be shown in multiple forms.

Point Form:

$(- 1, 1)$

Equation Form:

$x = - 1, y = 1$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.