Solve by Addition/Elimination 2a+3b=12 5a-b=13

$2 a + 3 b = 12$ $5 a - b = 13$

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

$2 a + 3 b = 12$

$(3) \cdot (5 a - b) = (3) (13)$

Simplify.

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Simplify $(3) \cdot (5 a - b)$ .

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Apply the distributive property.

$2 a + 3 b = 12$

$3 (5 a) + 3 (- b) = (3) (13)$

Multiply.

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Multiply $5$ by $3$ .

$2 a + 3 b = 12$

$15 a + 3 (- b) = (3) (13)$

Multiply $- 1$ by $3$ .

$2 a + 3 b = 12$

$15 a - 3 b = (3) (13)$

$2 a + 3 b = 12$

$15 a - 3 b = (3) (13)$

$2 a + 3 b = 12$

$15 a - 3 b = (3) (13)$

Multiply $3$ by $13$ .

$2 a + 3 b = 12$

$15 a - 3 b = 39$

$2 a + 3 b = 12$

$15 a - 3 b = 39$

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

		$2$	$a$	$+$	$3$	$b$	$=$	$1$	$2$
$+$	$1$	$5$	$a$	$-$	$3$	$b$	$=$	$3$	$9$
	$1$	$7$	$a$				$=$	$5$	$1$

Divide each term by $17$ and simplify.

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Divide each term in $17 a = 51$ by $17$ .

$\frac{17 a}{17} = \frac{51}{17}$

Cancel the common factor of $17$ .

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Cancel the common factor.

$\frac{17 a}{17} = \frac{51}{17}$

Divide $a$ by $1$ .

$a = \frac{51}{17}$

Divide $51$ by $17$ .

$a = 3$

Substitute the value found for $a$ into one of the original equations, then solve for $b$ .

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Substitute the value found for $a$ into one of the original equations to solve for $b$ .

$2 (3) + 3 b = 12$

Multiply $2$ by $3$ .

$6 + 3 b = 12$

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

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Subtract $6$ from both sides of the equation.

$3 b = 12 - 6$

Subtract $6$ from $12$ .

$3 b = 6$

Divide each term by $3$ and simplify.

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Divide each term in $3 b = 6$ by $3$ .

$\frac{3 b}{3} = \frac{6}{3}$

Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 b}{3} = \frac{6}{3}$

Divide $b$ by $1$ .

$b = \frac{6}{3}$

Divide $6$ by $3$ .

$b = 2$

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

$a = 3$

$b = 2$

Solve by Addition/Elimination 2a+3b=12 5a-b=13

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