How do you solve $\frac{b + 6}{b - 3} = 2$ $\frac { b + 6} { b - 3} = 2$ ?

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Answer 1 · 2016-11-15T02:59:17

$b = 12$ $b=12$

Multiply both sides by $b - 3$ $b-3$ .

$b + 6 = 2 (b - 3)$ $b+6=2(b-3)$

Distribute the 2

$b + 6 = 2 b - 6$ $b+6=2b-6$

Add $6$ $6$ to both sides

$b + 12 = 2 b$ $b+12=2b$

Subtract $b$ $b$ from both sides

$12 = b$ $12=b$

How do you solve b+6b−3=2\frac { b + 6} { b - 3} = 2?