Solve by Addition/Elimination 9a-3b=3 -3a+b=-1

$9 a - 3 b = 3$ $- 3 a + b = - 1$

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

$9 a - 3 b = 3$

$(3) \cdot (- 3 a + b) = (3) (- 1)$

Simplify.

Tap for more steps…

Simplify $(3) \cdot (- 3 a + b)$ .

Tap for more steps…

Apply the distributive property.

$9 a - 3 b = 3$

$3 (- 3 a) + 3 b = (3) (- 1)$

Multiply $- 3$ by $3$ .

$9 a - 3 b = 3$

$- 9 a + 3 b = (3) (- 1)$

$9 a - 3 b = 3$

$- 9 a + 3 b = (3) (- 1)$

Multiply $3$ by $- 1$ .

$9 a - 3 b = 3$

$- 9 a + 3 b = - 3$

$9 a - 3 b = 3$

$- 9 a + 3 b = - 3$

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

		$9$	$a$	$-$	$3$	$b$	$=$		$3$
$+$	$-$	$9$	$a$	$+$	$3$	$b$	$=$	$-$	$3$
						$0$	$=$		$0$

Since $0 = 0$ , the equations intersect at an infinite number of points.

Infinite number of solutions

Solve one of the equations for $b$ .

Tap for more steps…

Subtract $9 a$ from both sides of the equation.

$- 3 b = 3 - 9 a$

Divide each term by $- 3$ and simplify.

Tap for more steps…

Divide each term in $- 3 b = 3 - 9 a$ by $- 3$ .

$\frac{- 3 b}{- 3} = \frac{3}{- 3} + \frac{- 9 a}{- 3}$

Cancel the common factor of $- 3$ .

Tap for more steps…

Cancel the common factor.

$\frac{- 3 b}{- 3} = \frac{3}{- 3} + \frac{- 9 a}{- 3}$

Divide $b$ by $1$ .

$b = \frac{3}{- 3} + \frac{- 9 a}{- 3}$

Simplify each term.

Tap for more steps…

Divide $3$ by $- 3$ .

$b = - 1 + \frac{- 9 a}{- 3}$

Cancel the common factor of $- 9$ and $- 3$ .

Tap for more steps…

Factor $- 3$ out of $- 9 a$ .

$b = - 1 + \frac{- 3 (3 a)}{- 3}$

Cancel the common factors.

Tap for more steps…

Factor $- 3$ out of $- 3$ .

$b = - 1 + \frac{- 3 (3 a)}{- 3 (1)}$

Cancel the common factor.

$b = - 1 + \frac{- 3 (3 a)}{- 3 \cdot 1}$

Rewrite the expression.

$b = - 1 + \frac{3 a}{1}$

Divide $3 a$ by $1$ .

$b = - 1 + 3 a$

The solution is the set of ordered pairs that make $b = - 1 + 3 a$ true.

$(- 1 + 3 a, a)$

Solve by Addition/Elimination 9a-3b=3 -3a+b=-1

Need help with MATH HOMEWORK