Solve by Addition/Elimination 3x+5y=6 4x+2y=-6

$3 x + 5 y = 6$ $4 x + 2 y = - 6$

Step 1

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

$(- 2) \cdot (3 x + 5 y) = (- 2) (6)$

$(5) \cdot (4 x + 2 y) = (5) (- 6)$

Step 2

Simplify.

Simplify the left side.

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

Apply the distributive property.

$- 2 (3 x) - 2 (5 y) = (- 2) (6)$

$(5) \cdot (4 x + 2 y) = (5) (- 6)$

Multiply.

Multiply $3$ by $- 2$ .

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

$(5) \cdot (4 x + 2 y) = (5) (- 6)$

Multiply $5$ by $- 2$ .

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

$(5) \cdot (4 x + 2 y) = (5) (- 6)$

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

$(5) \cdot (4 x + 2 y) = (5) (- 6)$

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

$(5) \cdot (4 x + 2 y) = (5) (- 6)$

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

$(5) \cdot (4 x + 2 y) = (5) (- 6)$

Simplify the right side.

Multiply $- 2$ by $6$ .

$- 6 x - 10 y = - 12$

$(5) \cdot (4 x + 2 y) = (5) (- 6)$

$- 6 x - 10 y = - 12$

$(5) \cdot (4 x + 2 y) = (5) (- 6)$

Simplify the left side.

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

Apply the distributive property.

$- 6 x - 10 y = - 12$

$5 (4 x) + 5 (2 y) = (5) (- 6)$

Multiply.

Multiply $4$ by $5$ .

$- 6 x - 10 y = - 12$

$20 x + 5 (2 y) = (5) (- 6)$

Multiply $2$ by $5$ .

$- 6 x - 10 y = - 12$

$20 x + 10 y = (5) (- 6)$

$- 6 x - 10 y = - 12$

$20 x + 10 y = (5) (- 6)$

$- 6 x - 10 y = - 12$

$20 x + 10 y = (5) (- 6)$

$- 6 x - 10 y = - 12$

$20 x + 10 y = (5) (- 6)$

Simplify the right side.

Multiply $5$ by $- 6$ .

$- 6 x - 10 y = - 12$

$20 x + 10 y = - 30$

$- 6 x - 10 y = - 12$

$20 x + 10 y = - 30$

$- 6 x - 10 y = - 12$

$20 x + 10 y = - 30$

Step 3

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

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

Step 4

Divide each term in $14 x = - 42$ by $14$ and simplify.

Divide each term in $14 x = - 42$ by $14$ .

$\frac{14 x}{14} = \frac{- 42}{14}$

Simplify the left side.

Cancel the common factor of $14$ .

Cancel the common factor.

$\frac{14 x}{14} = \frac{- 42}{14}$

Divide $x$ by $1$ .

$x = \frac{- 42}{14}$

Simplify the right side.

Divide $- 42$ by $14$ .

$x = - 3$

Step 5

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 $y$ .

$- 6 \cdot - 3 - 10 y = - 12$

Multiply $- 6$ by $- 3$ .

$18 - 10 y = - 12$

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

Subtract $18$ from both sides of the equation.

$- 10 y = - 12 - 18$

Subtract $18$ from $- 12$ .

$- 10 y = - 30$

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 6

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

$(- 3, 3)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(- 3, 3)$

Equation Form:

$x = - 3, y = 3$

Step 8

Solve by Addition/Elimination 3x+5y=6 4x+2y=-6

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