Solve by Addition/Elimination 3x-5y=-39 3y+7x=41

$3 x - 5 y = - 39$ $3 y + 7 x = 41$

Step 1

Simplify the left side.

Reorder $3 y$ and $7 x$ .

$7 x + 3 y = 41$

$3 x - 5 y = - 39$

$7 x + 3 y = 41$

$3 x - 5 y = - 39$

Step 2

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

$(3) \cdot (3 x - 5 y) = (3) (- 39)$

$(5) \cdot (7 x + 3 y) = (5) (41)$

Step 3

Simplify.

Simplify the left side.

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

Apply the distributive property.

$3 (3 x) + 3 (- 5 y) = (3) (- 39)$

$(5) \cdot (7 x + 3 y) = (5) (41)$

Multiply.

Multiply $3$ by $3$ .

$9 x + 3 (- 5 y) = (3) (- 39)$

$(5) \cdot (7 x + 3 y) = (5) (41)$

Multiply $- 5$ by $3$ .

$9 x - 15 y = (3) (- 39)$

$(5) \cdot (7 x + 3 y) = (5) (41)$

$9 x - 15 y = (3) (- 39)$

$(5) \cdot (7 x + 3 y) = (5) (41)$

$9 x - 15 y = (3) (- 39)$

$(5) \cdot (7 x + 3 y) = (5) (41)$

$9 x - 15 y = (3) (- 39)$

$(5) \cdot (7 x + 3 y) = (5) (41)$

Simplify the right side.

Multiply $3$ by $- 39$ .

$9 x - 15 y = - 117$

$(5) \cdot (7 x + 3 y) = (5) (41)$

$9 x - 15 y = - 117$

$(5) \cdot (7 x + 3 y) = (5) (41)$

Simplify the left side.

Simplify $(5) \cdot (7 x + 3 y)$ .

Apply the distributive property.

$9 x - 15 y = - 117$

$5 (7 x) + 5 (3 y) = (5) (41)$

Multiply.

Multiply $7$ by $5$ .

$9 x - 15 y = - 117$

$35 x + 5 (3 y) = (5) (41)$

Multiply $3$ by $5$ .

$9 x - 15 y = - 117$

$35 x + 15 y = (5) (41)$

$9 x - 15 y = - 117$

$35 x + 15 y = (5) (41)$

$9 x - 15 y = - 117$

$35 x + 15 y = (5) (41)$

$9 x - 15 y = - 117$

$35 x + 15 y = (5) (41)$

Simplify the right side.

Multiply $5$ by $41$ .

$9 x - 15 y = - 117$

$35 x + 15 y = 205$

$9 x - 15 y = - 117$

$35 x + 15 y = 205$

$9 x - 15 y = - 117$

$35 x + 15 y = 205$

Step 4

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

		$9$	$x$	$-$	$1$	$5$	$y$	$=$	$-$	$1$	$1$	$7$
$+$	$3$	$5$	$x$	$+$	$1$	$5$	$y$	$=$		$2$	$0$	$5$
	$4$	$4$	$x$					$=$			$8$	$8$

Step 5

Divide each term in $44 x = 88$ by $44$ and simplify.

Divide each term in $44 x = 88$ by $44$ .

$\frac{44 x}{44} = \frac{88}{44}$

Simplify the left side.

Cancel the common factor of $44$ .

Cancel the common factor.

$\frac{44 x}{44} = \frac{88}{44}$

Divide $x$ by $1$ .

$x = \frac{88}{44}$

Simplify the right side.

Divide $88$ by $44$ .

$x = 2$

Step 6

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 $y$ .

$9 (2) - 15 y = - 117$

Multiply $9$ by $2$ .

$18 - 15 y = - 117$

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

Subtract $18$ from both sides of the equation.

$- 15 y = - 117 - 18$

Subtract $18$ from $- 117$ .

$- 15 y = - 135$

Divide each term in $- 15 y = - 135$ by $- 15$ and simplify.

Divide each term in $- 15 y = - 135$ by $- 15$ .

$\frac{- 15 y}{- 15} = \frac{- 135}{- 15}$

Simplify the left side.

Cancel the common factor of $- 15$ .

Cancel the common factor.

$\frac{- 15 y}{- 15} = \frac{- 135}{- 15}$

Divide $y$ by $1$ .

$y = \frac{- 135}{- 15}$

Simplify the right side.

Divide $- 135$ by $- 15$ .

$y = 9$

Step 7

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

$(2, 9)$

Step 8

The result can be shown in multiple forms.

Point Form:

$(2, 9)$

Equation Form:

$x = 2, y = 9$

Step 9

Solve by Addition/Elimination 3x-5y=-39 3y+7x=41

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