Solve by Addition/Elimination y-x=2 y=-14x+7

$y - x = 2$ $y = - 14 x + 7$

Simplify the left side.

Reorder $y$ and $- x$ .

$- x + y = 2$

$y = - 14 x + 7$

$- x + y = 2$

$y = - 14 x + 7$

Move all terms containing variables to the left side of the equation.

Add $14 x$ to both sides of the equation.

$y + 14 x = 7$

$- x + y = 2$

Reorder $y$ and $14 x$ .

$14 x + y = 7$

$- x + y = 2$

$14 x + y = 7$

$- x + y = 2$

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

$- x + y = 2$

$(- 1) \cdot (14 x + y) = (- 1) (7)$

Simplify.

Simplify the left side.

Simplify $(- 1) \cdot (14 x + y)$ .

Apply the distributive property.

$- x + y = 2$

$- 1 (14 x) - 1 y = (- 1) (7)$

Simplify the expression.

Multiply $14$ by $- 1$ .

$- x + y = 2$

$- 14 x - 1 y = (- 1) (7)$

Rewrite $- 1 y$ as $- y$ .

$- x + y = 2$

$- 14 x - y = (- 1) (7)$

$- x + y = 2$

$- 14 x - y = (- 1) (7)$

$- x + y = 2$

$- 14 x - y = (- 1) (7)$

$- x + y = 2$

$- 14 x - y = (- 1) (7)$

Simplify the right side.

Multiply $- 1$ by $7$ .

$- x + y = 2$

$- 14 x - y = - 7$

$- x + y = 2$

$- 14 x - y = - 7$

$- x + y = 2$

$- 14 x - y = - 7$

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

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

Divide each term in $- 15 x = - 5$ by $- 15$ and simplify.

Divide each term in $- 15 x = - 5$ by $- 15$ .

$\frac{- 15 x}{- 15} = \frac{- 5}{- 15}$

Simplify the left side.

Cancel the common factor of $- 15$ .

Cancel the common factor.

$\frac{- 15 x}{- 15} = \frac{- 5}{- 15}$

Divide $x$ by $1$ .

$x = \frac{- 5}{- 15}$

Simplify the right side.

Cancel the common factor of $- 5$ and $- 15$ .

Factor $- 5$ out of $- 5$ .

$x = \frac{- 5 (1)}{- 15}$

Cancel the common factors.

Factor $- 5$ out of $- 15$ .

$x = \frac{- 5 \cdot 1}{- 5 \cdot 3}$

Cancel the common factor.

$x = \frac{- 5 \cdot 1}{- 5 \cdot 3}$

Rewrite the expression.

$x = \frac{1}{3}$

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 $y$ .

$- (\frac{1}{3}) + y = 2$

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

Add $\frac{1}{3}$ to both sides of the equation.

$y = 2 + \frac{1}{3}$

To write $2$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = 2 \cdot \frac{3}{3} + \frac{1}{3}$

Combine $2$ and $\frac{3}{3}$ .

$y = \frac{2 \cdot 3}{3} + \frac{1}{3}$

Combine the numerators over the common denominator.

$y = \frac{2 \cdot 3 + 1}{3}$

Simplify the numerator.

Multiply $2$ by $3$ .

$y = \frac{6 + 1}{3}$

Add $6$ and $1$ .

$y = \frac{7}{3}$

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

$(\frac{1}{3}, \frac{7}{3})$

The result can be shown in multiple forms.

Point Form:

$(\frac{1}{3}, \frac{7}{3})$

Equation Form:

$x = \frac{1}{3}, y = \frac{7}{3}$

Solve by Addition/Elimination y-x=2 y=-14x+7

Related Questions

Support LCMGCF.com by disabling your adblocker.