Solve by Addition/Elimination x=5y , 2x-3y=14

$x = 5 y$ , $2 x - 3 y = 14$

Subtract $5 y$ from both sides of the equation.

$x - 5 y = 0$

$2 x - 3 y = 14$

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

$(- 2) \cdot (x - 5 y) = (- 2) (0)$

$2 x - 3 y = 14$

Simplify.

Simplify the left side.

Simplify $(- 2) \cdot (x - 5 y)$ .

Apply the distributive property.

$- 2 x - 2 (- 5 y) = (- 2) (0)$

$2 x - 3 y = 14$

Multiply $- 5$ by $- 2$ .

$- 2 x + 10 y = (- 2) (0)$

$2 x - 3 y = 14$

$- 2 x + 10 y = (- 2) (0)$

$2 x - 3 y = 14$

$- 2 x + 10 y = (- 2) (0)$

$2 x - 3 y = 14$

Simplify the right side.

Multiply $- 2$ by $0$ .

$- 2 x + 10 y = 0$

$2 x - 3 y = 14$

$- 2 x + 10 y = 0$

$2 x - 3 y = 14$

$- 2 x + 10 y = 0$

$2 x - 3 y = 14$

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

	$-$	$2$	$x$	$+$	$1$	$0$	$y$	$=$		$0$
$+$		$2$	$x$		$-$	$3$	$y$	$=$	$1$	$4$
						$7$	$y$	$=$	$1$	$4$

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

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

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

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Divide $14$ by $7$ .

$y = 2$

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 $x$ .

$- 2 x + 10 (2) = 0$

Multiply $10$ by $2$ .

$- 2 x + 20 = 0$

Subtract $20$ from both sides of the equation.

$- 2 x = - 20$

Divide each term in $- 2 x = - 20$ by $- 2$ and simplify.

Divide each term in $- 2 x = - 20$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{- 20}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{- 20}{- 2}$

Divide $x$ by $1$ .

$x = \frac{- 20}{- 2}$

Simplify the right side.

Divide $- 20$ by $- 2$ .

$x = 10$

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

$(10, 2)$

The result can be shown in multiple forms.

Point Form:

$(10, 2)$

Equation Form:

$x = 10, y = 2$

Solve by Addition/Elimination x=5y , 2x-3y=14

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