Solve by Addition/Elimination 4x-2y+7=33 5y=3x-30

$4 x - 2 y + 7 = 33$ $5 y = 3 x - 30$

Step 1

Move all terms containing variables to the left.

Subtract $7$ from both sides of the equation.

$4 x - 2 y = 33 - 7, 5 y = 3 x - 30$

Subtract $7$ from $33$ .

$4 x - 2 y = 26, 5 y = 3 x - 30$

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

$4 x - 2 y = 26, 5 y - 3 x = - 30$

Step 2

Reorder the polynomial.

$- 3 x + 5 y = - 30$

$4 x - 2 y = 26$

Step 3

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

$(5) \cdot (4 x - 2 y) = (5) (26)$

$(2) \cdot (- 3 x + 5 y) = (2) (- 30)$

Step 4

Simplify.

Simplify the left side.

Simplify $(5) \cdot (4 x - 2 y)$ .

Apply the distributive property.

$5 (4 x) + 5 (- 2 y) = (5) (26)$

$(2) \cdot (- 3 x + 5 y) = (2) (- 30)$

Multiply.

Multiply $4$ by $5$ .

$20 x + 5 (- 2 y) = (5) (26)$

$(2) \cdot (- 3 x + 5 y) = (2) (- 30)$

Multiply $- 2$ by $5$ .

$20 x - 10 y = (5) (26)$

$(2) \cdot (- 3 x + 5 y) = (2) (- 30)$

$20 x - 10 y = (5) (26)$

$(2) \cdot (- 3 x + 5 y) = (2) (- 30)$

$20 x - 10 y = (5) (26)$

$(2) \cdot (- 3 x + 5 y) = (2) (- 30)$

$20 x - 10 y = (5) (26)$

$(2) \cdot (- 3 x + 5 y) = (2) (- 30)$

Simplify the right side.

Multiply $5$ by $26$ .

$20 x - 10 y = 130$

$(2) \cdot (- 3 x + 5 y) = (2) (- 30)$

$20 x - 10 y = 130$

$(2) \cdot (- 3 x + 5 y) = (2) (- 30)$

Simplify the left side.

Simplify $(2) \cdot (- 3 x + 5 y)$ .

Apply the distributive property.

$20 x - 10 y = 130$

$2 (- 3 x) + 2 (5 y) = (2) (- 30)$

Multiply.

Multiply $- 3$ by $2$ .

$20 x - 10 y = 130$

$- 6 x + 2 (5 y) = (2) (- 30)$

Multiply $5$ by $2$ .

$20 x - 10 y = 130$

$- 6 x + 10 y = (2) (- 30)$

$20 x - 10 y = 130$

$- 6 x + 10 y = (2) (- 30)$

$20 x - 10 y = 130$

$- 6 x + 10 y = (2) (- 30)$

$20 x - 10 y = 130$

$- 6 x + 10 y = (2) (- 30)$

Simplify the right side.

Multiply $2$ by $- 30$ .

$20 x - 10 y = 130$

$- 6 x + 10 y = - 60$

$20 x - 10 y = 130$

$- 6 x + 10 y = - 60$

$20 x - 10 y = 130$

$- 6 x + 10 y = - 60$

Step 5

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

	$2$	$0$	$x$	$-$	$1$	$0$	$y$	$=$	$1$	$3$	$0$
$+$	$-$	$6$	$x$	$+$	$1$	$0$	$y$	$=$	$-$	$6$	$0$
	$1$	$4$	$x$					$=$		$7$	$0$

Step 6

Divide each term in $14 x = 70$ by $14$ and simplify.

Divide each term in $14 x = 70$ by $14$ .

$\frac{14 x}{14} = \frac{70}{14}$

Simplify the left side.

Cancel the common factor of $14$ .

Cancel the common factor.

$\frac{14 x}{14} = \frac{70}{14}$

Divide $x$ by $1$ .

$x = \frac{70}{14}$

Simplify the right side.

Divide $70$ by $14$ .

$x = 5$

Step 7

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 $y$ .

$20 (5) - 10 y = 130$

Multiply $20$ by $5$ .

$100 - 10 y = 130$

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

Subtract $100$ from both sides of the equation.

$- 10 y = 130 - 100$

Subtract $100$ from $130$ .

$- 10 y = 30$

Divide each term in $- 10 y = 30$ by $- 10$ and simplify.

Divide each term in $- 10 y = 30$ by $- 10$ .

$\frac{- 10 y}{- 10} = \frac{30}{- 10}$

Simplify the left side.

Cancel the common factor of $- 10$ .

Cancel the common factor.

$\frac{- 10 y}{- 10} = \frac{30}{- 10}$

Divide $y$ by $1$ .

$y = \frac{30}{- 10}$

Simplify the right side.

Divide $30$ by $- 10$ .

$y = - 3$

Step 8

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

$(5, - 3)$

Step 9

The result can be shown in multiple forms.

Point Form:

$(5, - 3)$

Equation Form:

$x = 5, y = - 3$

Step 10

Solve by Addition/Elimination 4x-2y+7=33 5y=3x-30

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