Solve by Addition/Elimination 5x-3y=23 7x-4y=33

$5 x - 3 y = 23$ $7 x - 4 y = 33$

Step 1

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

$(- 4) \cdot (5 x - 3 y) = (- 4) (23)$

$(3) \cdot (7 x - 4 y) = (3) (33)$

Step 2

Simplify.

Simplify the left side.

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

Apply the distributive property.

$- 4 (5 x) - 4 (- 3 y) = (- 4) (23)$

$(3) \cdot (7 x - 4 y) = (3) (33)$

Multiply.

Multiply $5$ by $- 4$ .

$- 20 x - 4 (- 3 y) = (- 4) (23)$

$(3) \cdot (7 x - 4 y) = (3) (33)$

Multiply $- 3$ by $- 4$ .

$- 20 x + 12 y = (- 4) (23)$

$(3) \cdot (7 x - 4 y) = (3) (33)$

$- 20 x + 12 y = (- 4) (23)$

$(3) \cdot (7 x - 4 y) = (3) (33)$

$- 20 x + 12 y = (- 4) (23)$

$(3) \cdot (7 x - 4 y) = (3) (33)$

$- 20 x + 12 y = (- 4) (23)$

$(3) \cdot (7 x - 4 y) = (3) (33)$

Simplify the right side.

Multiply $- 4$ by $23$ .

$- 20 x + 12 y = - 92$

$(3) \cdot (7 x - 4 y) = (3) (33)$

$- 20 x + 12 y = - 92$

$(3) \cdot (7 x - 4 y) = (3) (33)$

Simplify the left side.

Simplify $(3) \cdot (7 x - 4 y)$ .

Apply the distributive property.

$- 20 x + 12 y = - 92$

$3 (7 x) + 3 (- 4 y) = (3) (33)$

Multiply.

Multiply $7$ by $3$ .

$- 20 x + 12 y = - 92$

$21 x + 3 (- 4 y) = (3) (33)$

Multiply $- 4$ by $3$ .

$- 20 x + 12 y = - 92$

$21 x - 12 y = (3) (33)$

$- 20 x + 12 y = - 92$

$21 x - 12 y = (3) (33)$

$- 20 x + 12 y = - 92$

$21 x - 12 y = (3) (33)$

$- 20 x + 12 y = - 92$

$21 x - 12 y = (3) (33)$

Simplify the right side.

Multiply $3$ by $33$ .

$- 20 x + 12 y = - 92$

$21 x - 12 y = 99$

$- 20 x + 12 y = - 92$

$21 x - 12 y = 99$

$- 20 x + 12 y = - 92$

$21 x - 12 y = 99$

Step 3

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

	$-$	$2$	$0$	$x$	$+$	$1$	$2$	$y$	$=$	$-$	$9$	$2$
$+$		$2$	$1$	$x$	$-$	$1$	$2$	$y$	$=$		$9$	$9$
				$x$					$=$			$7$

Step 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 $y$ .

$- 20 \cdot 7 + 12 y = - 92$

Multiply $- 20$ by $7$ .

$- 140 + 12 y = - 92$

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

Add $140$ to both sides of the equation.

$12 y = - 92 + 140$

Add $- 92$ and $140$ .

$12 y = 48$

Divide each term in $12 y = 48$ by $12$ and simplify.

Divide each term in $12 y = 48$ by $12$ .

$\frac{12 y}{12} = \frac{48}{12}$

Simplify the left side.

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 y}{12} = \frac{48}{12}$

Divide $y$ by $1$ .

$y = \frac{48}{12}$

Simplify the right side.

Divide $48$ by $12$ .

$y = 4$

Step 5

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

$(7, 4)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(7, 4)$

Equation Form:

$x = 7, y = 4$

Step 7

Solve by Addition/Elimination 5x-3y=23 7x-4y=33

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