Solve by Addition/Elimination 4x+3y=-20 , 6x+12y=-10

$4 x + 3 y = - 20$ , $6 x + 12 y = - 10$

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

$(- 3) \cdot (4 x + 3 y) = (- 3) (- 20)$

$(2) \cdot (6 x + 12 y) = (2) (- 10)$

Simplify.

Simplify the left side.

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

Apply the distributive property.

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

$(2) \cdot (6 x + 12 y) = (2) (- 10)$

Multiply.

Multiply $4$ by $- 3$ .

$- 12 x - 3 (3 y) = (- 3) (- 20)$

$(2) \cdot (6 x + 12 y) = (2) (- 10)$

Multiply $3$ by $- 3$ .

$- 12 x - 9 y = (- 3) (- 20)$

$(2) \cdot (6 x + 12 y) = (2) (- 10)$

$- 12 x - 9 y = (- 3) (- 20)$

$(2) \cdot (6 x + 12 y) = (2) (- 10)$

$- 12 x - 9 y = (- 3) (- 20)$

$(2) \cdot (6 x + 12 y) = (2) (- 10)$

$- 12 x - 9 y = (- 3) (- 20)$

$(2) \cdot (6 x + 12 y) = (2) (- 10)$

Simplify the right side.

Multiply $- 3$ by $- 20$ .

$- 12 x - 9 y = 60$

$(2) \cdot (6 x + 12 y) = (2) (- 10)$

$- 12 x - 9 y = 60$

$(2) \cdot (6 x + 12 y) = (2) (- 10)$

Simplify the left side.

Simplify $(2) \cdot (6 x + 12 y)$ .

Apply the distributive property.

$- 12 x - 9 y = 60$

$2 (6 x) + 2 (12 y) = (2) (- 10)$

Multiply.

Multiply $6$ by $2$ .

$- 12 x - 9 y = 60$

$12 x + 2 (12 y) = (2) (- 10)$

Multiply $12$ by $2$ .

$- 12 x - 9 y = 60$

$12 x + 24 y = (2) (- 10)$

$- 12 x - 9 y = 60$

$12 x + 24 y = (2) (- 10)$

$- 12 x - 9 y = 60$

$12 x + 24 y = (2) (- 10)$

$- 12 x - 9 y = 60$

$12 x + 24 y = (2) (- 10)$

Simplify the right side.

Multiply $2$ by $- 10$ .

$- 12 x - 9 y = 60$

$12 x + 24 y = - 20$

$- 12 x - 9 y = 60$

$12 x + 24 y = - 20$

$- 12 x - 9 y = 60$

$12 x + 24 y = - 20$

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

	$-$	$1$	$2$	$x$		$-$	$9$	$y$	$=$		$6$	$0$
$+$		$1$	$2$	$x$	$+$	$2$	$4$	$y$	$=$	$-$	$2$	$0$
						$1$	$5$	$y$	$=$		$4$	$0$

Divide each term in $15 y = 40$ by $15$ and simplify.

Divide each term in $15 y = 40$ by $15$ .

$\frac{15 y}{15} = \frac{40}{15}$

Simplify the left side.

Cancel the common factor of $15$ .

Cancel the common factor.

$\frac{15 y}{15} = \frac{40}{15}$

Divide $y$ by $1$ .

$y = \frac{40}{15}$

Simplify the right side.

Cancel the common factor of $40$ and $15$ .

Factor $5$ out of $40$ .

$y = \frac{5 (8)}{15}$

Cancel the common factors.

Factor $5$ out of $15$ .

$y = \frac{5 \cdot 8}{5 \cdot 3}$

Cancel the common factor.

$y = \frac{5 \cdot 8}{5 \cdot 3}$

Rewrite the expression.

$y = \frac{8}{3}$

Substitute the value found for $y$ into one of the original equations, then solve for $x$ .

Substitute the value found for $y$ into one of the original equations to solve for $x$ .

$- 12 x - 9 (\frac{8}{3}) = 60$

Simplify each term.

Cancel the common factor of $3$ .

Factor $3$ out of $- 9$ .

$- 12 x + 3 (- 3) \frac{8}{3} = 60$

Cancel the common factor.

$- 12 x + 3 \cdot - 3 \frac{8}{3} = 60$

Rewrite the expression.

$- 12 x - 3 \cdot 8 = 60$

Multiply $- 3$ by $8$ .

$- 12 x - 24 = 60$

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

Add $24$ to both sides of the equation.

$- 12 x = 60 + 24$

Add $60$ and $24$ .

$- 12 x = 84$

Divide each term in $- 12 x = 84$ by $- 12$ and simplify.

Divide each term in $- 12 x = 84$ by $- 12$ .

$\frac{- 12 x}{- 12} = \frac{84}{- 12}$

Simplify the left side.

Cancel the common factor of $- 12$ .

Cancel the common factor.

$\frac{- 12 x}{- 12} = \frac{84}{- 12}$

Divide $x$ by $1$ .

$x = \frac{84}{- 12}$

Simplify the right side.

Divide $84$ by $- 12$ .

$x = - 7$

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

$(- 7, \frac{8}{3})$

The result can be shown in multiple forms.

Point Form:

$(- 7, \frac{8}{3})$

Equation Form:

$x = - 7, y = \frac{8}{3}$

Solve by Addition/Elimination 4x+3y=-20 , 6x+12y=-10

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