Solve by Addition/Elimination -5y+6x=40 3y-8x=-46

$- 5 y + 6 x = 40$ $3 y - 8 x = - 46$

Step 1

Simplify the left side.

Reorder $- 5 y$ and $6 x$ .

$6 x - 5 y = 40$

$3 y - 8 x = - 46$

$6 x - 5 y = 40$

$3 y - 8 x = - 46$

Step 2

Simplify the left side.

Reorder $3 y$ and $- 8 x$ .

$- 8 x + 3 y = - 46$

$6 x - 5 y = 40$

$- 8 x + 3 y = - 46$

$6 x - 5 y = 40$

Step 3

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

$(3) \cdot (6 x - 5 y) = (3) (40)$

$(5) \cdot (- 8 x + 3 y) = (5) (- 46)$

Step 4

Simplify.

Simplify the left side.

Simplify $(3) \cdot (6 x - 5 y)$ .

Apply the distributive property.

$3 (6 x) + 3 (- 5 y) = (3) (40)$

$(5) \cdot (- 8 x + 3 y) = (5) (- 46)$

Multiply.

Multiply $6$ by $3$ .

$18 x + 3 (- 5 y) = (3) (40)$

$(5) \cdot (- 8 x + 3 y) = (5) (- 46)$

Multiply $- 5$ by $3$ .

$18 x - 15 y = (3) (40)$

$(5) \cdot (- 8 x + 3 y) = (5) (- 46)$

$18 x - 15 y = (3) (40)$

$(5) \cdot (- 8 x + 3 y) = (5) (- 46)$

$18 x - 15 y = (3) (40)$

$(5) \cdot (- 8 x + 3 y) = (5) (- 46)$

$18 x - 15 y = (3) (40)$

$(5) \cdot (- 8 x + 3 y) = (5) (- 46)$

Simplify the right side.

Multiply $3$ by $40$ .

$18 x - 15 y = 120$

$(5) \cdot (- 8 x + 3 y) = (5) (- 46)$

$18 x - 15 y = 120$

$(5) \cdot (- 8 x + 3 y) = (5) (- 46)$

Simplify the left side.

Simplify $(5) \cdot (- 8 x + 3 y)$ .

Apply the distributive property.

$18 x - 15 y = 120$

$5 (- 8 x) + 5 (3 y) = (5) (- 46)$

Multiply.

Multiply $- 8$ by $5$ .

$18 x - 15 y = 120$

$- 40 x + 5 (3 y) = (5) (- 46)$

Multiply $3$ by $5$ .

$18 x - 15 y = 120$

$- 40 x + 15 y = (5) (- 46)$

$18 x - 15 y = 120$

$- 40 x + 15 y = (5) (- 46)$

$18 x - 15 y = 120$

$- 40 x + 15 y = (5) (- 46)$

$18 x - 15 y = 120$

$- 40 x + 15 y = (5) (- 46)$

Simplify the right side.

Multiply $5$ by $- 46$ .

$18 x - 15 y = 120$

$- 40 x + 15 y = - 230$

$18 x - 15 y = 120$

$- 40 x + 15 y = - 230$

$18 x - 15 y = 120$

$- 40 x + 15 y = - 230$

Step 5

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

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

Step 6

Divide each term in $- 22 x = - 110$ by $- 22$ and simplify.

Divide each term in $- 22 x = - 110$ by $- 22$ .

$\frac{- 22 x}{- 22} = \frac{- 110}{- 22}$

Simplify the left side.

Cancel the common factor of $- 22$ .

Cancel the common factor.

$\frac{- 22 x}{- 22} = \frac{- 110}{- 22}$

Divide $x$ by $1$ .

$x = \frac{- 110}{- 22}$

Simplify the right side.

Divide $- 110$ by $- 22$ .

$x = 5$

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 $y$ .

$18 (5) - 15 y = 120$

Multiply $18$ by $5$ .

$90 - 15 y = 120$

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

Subtract $90$ from both sides of the equation.

$- 15 y = 120 - 90$

Subtract $90$ from $120$ .

$- 15 y = 30$

Divide each term in $- 15 y = 30$ by $- 15$ and simplify.

Divide each term in $- 15 y = 30$ by $- 15$ .

$\frac{- 15 y}{- 15} = \frac{30}{- 15}$

Simplify the left side.

Cancel the common factor of $- 15$ .

Cancel the common factor.

$\frac{- 15 y}{- 15} = \frac{30}{- 15}$

Divide $y$ by $1$ .

$y = \frac{30}{- 15}$

Simplify the right side.

Divide $30$ by $- 15$ .

$y = - 2$

Step 8

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

$(5, - 2)$

Step 9

The result can be shown in multiple forms.

Point Form:

$(5, - 2)$

Equation Form:

$x = 5, y = - 2$

Step 10

Solve by Addition/Elimination -5y+6x=40 3y-8x=-46

Related Questions

Support LCMGCF.com by disabling your adblocker.