Solve by Addition/Elimination 5x=4y+8 3y=3x-3

$5 x = 4 y + 8$ $3 y = 3 x - 3$

Subtract $4 y$ from both sides of the equation.

$5 x - 4 y = 8$

$3 y = 3 x - 3$

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

Subtract $3 x$ from both sides of the equation.

$3 y - 3 x = - 3$

$5 x - 4 y = 8$

Reorder $3 y$ and $- 3 x$ .

$- 3 x + 3 y = - 3$

$5 x - 4 y = 8$

$- 3 x + 3 y = - 3$

$5 x - 4 y = 8$

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

$(3) \cdot (5 x - 4 y) = (3) (8)$

$(4) \cdot (- 3 x + 3 y) = (4) (- 3)$

Simplify.

Simplify the left side.

Simplify $(3) \cdot (5 x - 4 y)$ .

Apply the distributive property.

$3 (5 x) + 3 (- 4 y) = (3) (8)$

$(4) \cdot (- 3 x + 3 y) = (4) (- 3)$

Multiply.

Multiply $5$ by $3$ .

$15 x + 3 (- 4 y) = (3) (8)$

$(4) \cdot (- 3 x + 3 y) = (4) (- 3)$

Multiply $- 4$ by $3$ .

$15 x - 12 y = (3) (8)$

$(4) \cdot (- 3 x + 3 y) = (4) (- 3)$

$15 x - 12 y = (3) (8)$

$(4) \cdot (- 3 x + 3 y) = (4) (- 3)$

$15 x - 12 y = (3) (8)$

$(4) \cdot (- 3 x + 3 y) = (4) (- 3)$

$15 x - 12 y = (3) (8)$

$(4) \cdot (- 3 x + 3 y) = (4) (- 3)$

Simplify the right side.

Multiply $3$ by $8$ .

$15 x - 12 y = 24$

$(4) \cdot (- 3 x + 3 y) = (4) (- 3)$

$15 x - 12 y = 24$

$(4) \cdot (- 3 x + 3 y) = (4) (- 3)$

Simplify the left side.

Simplify $(4) \cdot (- 3 x + 3 y)$ .

Apply the distributive property.

$15 x - 12 y = 24$

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

Multiply.

Multiply $- 3$ by $4$ .

$15 x - 12 y = 24$

$- 12 x + 4 (3 y) = (4) (- 3)$

Multiply $3$ by $4$ .

$15 x - 12 y = 24$

$- 12 x + 12 y = (4) (- 3)$

$15 x - 12 y = 24$

$- 12 x + 12 y = (4) (- 3)$

$15 x - 12 y = 24$

$- 12 x + 12 y = (4) (- 3)$

$15 x - 12 y = 24$

$- 12 x + 12 y = (4) (- 3)$

Simplify the right side.

Multiply $4$ by $- 3$ .

$15 x - 12 y = 24$

$- 12 x + 12 y = - 12$

$15 x - 12 y = 24$

$- 12 x + 12 y = - 12$

$15 x - 12 y = 24$

$- 12 x + 12 y = - 12$

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

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

Divide each term in $3 x = 12$ by $3$ and simplify.

Divide each term in $3 x = 12$ by $3$ .

$\frac{3 x}{3} = \frac{12}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{12}{3}$

Divide $x$ by $1$ .

$x = \frac{12}{3}$

$x = \frac{12}{3}$

$x = \frac{12}{3}$

Simplify the right side.

Divide $12$ by $3$ .

$x = 4$

$x = 4$

$x = 4$

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 $y$ .

$15 (4) - 12 y = 24$

Multiply $15$ by $4$ .

$60 - 12 y = 24$

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

Subtract $60$ from both sides of the equation.

$- 12 y = 24 - 60$

Subtract $60$ from $24$ .

$- 12 y = - 36$

$- 12 y = - 36$

Divide each term in $- 12 y = - 36$ by $- 12$ and simplify.

Divide each term in $- 12 y = - 36$ by $- 12$ .

$\frac{- 12 y}{- 12} = \frac{- 36}{- 12}$

Simplify the left side.

Cancel the common factor of $- 12$ .

Cancel the common factor.

$\frac{- 12 y}{- 12} = \frac{- 36}{- 12}$

Divide $y$ by $1$ .

$y = \frac{- 36}{- 12}$

$y = \frac{- 36}{- 12}$

$y = \frac{- 36}{- 12}$

Simplify the right side.

Divide $- 36$ by $- 12$ .

$y = 3$

$y = 3$

$y = 3$

$y = 3$

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

$(4, 3)$

The result can be shown in multiple forms.

Point Form:

$(4, 3)$

Equation Form:

$x = 4, y = 3$

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