Solve by Addition/Elimination 3x+2y=16 x+5y=14

$3 x + 2 y = 16$ $x + 5 y = 14$

Step 1

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

$3 x + 2 y = 16$

$(- 3) \cdot (x + 5 y) = (- 3) (14)$

Step 2

Simplify.

Simplify the left side.

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

Apply the distributive property.

$3 x + 2 y = 16$

$- 3 x - 3 (5 y) = (- 3) (14)$

Multiply $5$ by $- 3$ .

$3 x + 2 y = 16$

$- 3 x - 15 y = (- 3) (14)$

$3 x + 2 y = 16$

$- 3 x - 15 y = (- 3) (14)$

$3 x + 2 y = 16$

$- 3 x - 15 y = (- 3) (14)$

Simplify the right side.

Multiply $- 3$ by $14$ .

$3 x + 2 y = 16$

$- 3 x - 15 y = - 42$

$3 x + 2 y = 16$

$- 3 x - 15 y = - 42$

$3 x + 2 y = 16$

$- 3 x - 15 y = - 42$

Step 3

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

		$3$	$x$	$+$		$2$	$y$	$=$		$1$	$6$
$+$	$-$	$3$	$x$	$-$	$1$	$5$	$y$	$=$	$-$	$4$	$2$
				$-$	$1$	$3$	$y$	$=$	$-$	$2$	$6$

Step 4

Divide each term in $- 13 y = - 26$ by $- 13$ and simplify.

Divide each term in $- 13 y = - 26$ by $- 13$ .

$\frac{- 13 y}{- 13} = \frac{- 26}{- 13}$

Simplify the left side.

Cancel the common factor of $- 13$ .

Cancel the common factor.

$\frac{- 13 y}{- 13} = \frac{- 26}{- 13}$

Divide $y$ by $1$ .

$y = \frac{- 26}{- 13}$

Simplify the right side.

Divide $- 26$ by $- 13$ .

$y = 2$

Step 5

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 $x$ .

$3 x + 2 (2) = 16$

Multiply $2$ by $2$ .

$3 x + 4 = 16$

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

Subtract $4$ from both sides of the equation.

$3 x = 16 - 4$

Subtract $4$ from $16$ .

$3 x = 12$

Divide each term in $3 x = 12$ by $3$ and simplify.

Divide each term in $3 x = 12$ by $3$ .

$\frac{3 x}{3} = \frac{12}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{12}{3}$

Divide $x$ by $1$ .

$x = \frac{12}{3}$

Simplify the right side.

Divide $12$ by $3$ .

$x = 4$

Step 6

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

$(4, 2)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(4, 2)$

Equation Form:

$x = 4, y = 2$

Step 8

Solve by Addition/Elimination 3x+2y=16 x+5y=14

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