Solve by Addition/Elimination 3p+9q=6 5p-5q=30

$3 p + 9 q = 6$ $5 p - 5 q = 30$

Step 1

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

$(- 5) \cdot (3 p + 9 q) = (- 5) (6)$

$(3) \cdot (5 p - 5 q) = (3) (30)$

Step 2

Simplify.

Simplify the left side.

Simplify $(- 5) \cdot (3 p + 9 q)$ .

Apply the distributive property.

$- 5 (3 p) - 5 (9 q) = (- 5) (6)$

$(3) \cdot (5 p - 5 q) = (3) (30)$

Multiply.

Multiply $3$ by $- 5$ .

$- 15 p - 5 (9 q) = (- 5) (6)$

$(3) \cdot (5 p - 5 q) = (3) (30)$

Multiply $9$ by $- 5$ .

$- 15 p - 45 q = (- 5) (6)$

$(3) \cdot (5 p - 5 q) = (3) (30)$

$- 15 p - 45 q = (- 5) (6)$

$(3) \cdot (5 p - 5 q) = (3) (30)$

$- 15 p - 45 q = (- 5) (6)$

$(3) \cdot (5 p - 5 q) = (3) (30)$

$- 15 p - 45 q = (- 5) (6)$

$(3) \cdot (5 p - 5 q) = (3) (30)$

Simplify the right side.

Multiply $- 5$ by $6$ .

$- 15 p - 45 q = - 30$

$(3) \cdot (5 p - 5 q) = (3) (30)$

$- 15 p - 45 q = - 30$

$(3) \cdot (5 p - 5 q) = (3) (30)$

Simplify the left side.

Simplify $(3) \cdot (5 p - 5 q)$ .

Apply the distributive property.

$- 15 p - 45 q = - 30$

$3 (5 p) + 3 (- 5 q) = (3) (30)$

Multiply.

Multiply $5$ by $3$ .

$- 15 p - 45 q = - 30$

$15 p + 3 (- 5 q) = (3) (30)$

Multiply $- 5$ by $3$ .

$- 15 p - 45 q = - 30$

$15 p - 15 q = (3) (30)$

$- 15 p - 45 q = - 30$

$15 p - 15 q = (3) (30)$

$- 15 p - 45 q = - 30$

$15 p - 15 q = (3) (30)$

$- 15 p - 45 q = - 30$

$15 p - 15 q = (3) (30)$

Simplify the right side.

Multiply $3$ by $30$ .

$- 15 p - 45 q = - 30$

$15 p - 15 q = 90$

$- 15 p - 45 q = - 30$

$15 p - 15 q = 90$

$- 15 p - 45 q = - 30$

$15 p - 15 q = 90$

Step 3

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

	$-$	$1$	$5$	$p$	$-$	$4$	$5$	$q$	$=$	$-$	$3$	$0$
$+$		$1$	$5$	$p$	$-$	$1$	$5$	$q$	$=$		$9$	$0$
					$-$	$6$	$0$	$q$	$=$		$6$	$0$

Step 4

Divide each term in $- 60 q = 60$ by $- 60$ and simplify.

Divide each term in $- 60 q = 60$ by $- 60$ .

$\frac{- 60 q}{- 60} = \frac{60}{- 60}$

Simplify the left side.

Cancel the common factor of $- 60$ .

Cancel the common factor.

$\frac{- 60 q}{- 60} = \frac{60}{- 60}$

Divide $q$ by $1$ .

$q = \frac{60}{- 60}$

Simplify the right side.

Divide $60$ by $- 60$ .

$q = - 1$

Step 5

Substitute the value found for $q$ into one of the original equations, then solve for $p$ .

Substitute the value found for $q$ into one of the original equations to solve for $p$ .

$- 15 p - 45 \cdot - 1 = - 30$

Multiply $- 45$ by $- 1$ .

$- 15 p + 45 = - 30$

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

Subtract $45$ from both sides of the equation.

$- 15 p = - 30 - 45$

Subtract $45$ from $- 30$ .

$- 15 p = - 75$

Divide each term in $- 15 p = - 75$ by $- 15$ and simplify.

Divide each term in $- 15 p = - 75$ by $- 15$ .

$\frac{- 15 p}{- 15} = \frac{- 75}{- 15}$

Simplify the left side.

Cancel the common factor of $- 15$ .

Cancel the common factor.

$\frac{- 15 p}{- 15} = \frac{- 75}{- 15}$

Divide $p$ by $1$ .

$p = \frac{- 75}{- 15}$

Simplify the right side.

Divide $- 75$ by $- 15$ .

$p = 5$

Step 6

This is the final solution to the independent system of equations.

$q = - 1$

$p = 5$

Solve by Addition/Elimination 3p+9q=6 5p-5q=30

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