Solve by Factoring 3/(3b+4)=2/(b-4)

$\frac{3}{3 b + 4} = \frac{2}{b - 4}$

Step 1

Subtract $\frac{2}{b - 4}$ from both sides of the equation.

$\frac{3}{3 b + 4} - \frac{2}{b - 4} = 0$

Step 2

Simplify $\frac{3}{3 b + 4} - \frac{2}{b - 4}$ .

To write $\frac{3}{3 b + 4}$ as a fraction with a common denominator, multiply by $\frac{b - 4}{b - 4}$ .

$\frac{3}{3 b + 4} \cdot \frac{b - 4}{b - 4} - \frac{2}{b - 4} = 0$

To write $- \frac{2}{b - 4}$ as a fraction with a common denominator, multiply by $\frac{3 b + 4}{3 b + 4}$ .

$\frac{3}{3 b + 4} \cdot \frac{b - 4}{b - 4} - \frac{2}{b - 4} \cdot \frac{3 b + 4}{3 b + 4} = 0$

Write each expression with a common denominator of $(3 b + 4) (b - 4)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{3}{3 b + 4}$ by $\frac{b - 4}{b - 4}$ .

$\frac{3 (b - 4)}{(3 b + 4) (b - 4)} - \frac{2}{b - 4} \cdot \frac{3 b + 4}{3 b + 4} = 0$

Multiply $\frac{2}{b - 4}$ by $\frac{3 b + 4}{3 b + 4}$ .

$\frac{3 (b - 4)}{(3 b + 4) (b - 4)} - \frac{2 (3 b + 4)}{(b - 4) (3 b + 4)} = 0$

Reorder the factors of $(b - 4) (3 b + 4)$ .

$\frac{3 (b - 4)}{(3 b + 4) (b - 4)} - \frac{2 (3 b + 4)}{(3 b + 4) (b - 4)} = 0$

Combine the numerators over the common denominator.

$\frac{3 (b - 4) - 2 (3 b + 4)}{(3 b + 4) (b - 4)} = 0$

Simplify the numerator.

Apply the distributive property.

$\frac{3 b + 3 \cdot - 4 - 2 (3 b + 4)}{(3 b + 4) (b - 4)} = 0$

Multiply $3$ by $- 4$ .

$\frac{3 b - 12 - 2 (3 b + 4)}{(3 b + 4) (b - 4)} = 0$

Apply the distributive property.

$\frac{3 b - 12 - 2 (3 b) - 2 \cdot 4}{(3 b + 4) (b - 4)} = 0$

Multiply $3$ by $- 2$ .

$\frac{3 b - 12 - 6 b - 2 \cdot 4}{(3 b + 4) (b - 4)} = 0$

Multiply $- 2$ by $4$ .

$\frac{3 b - 12 - 6 b - 8}{(3 b + 4) (b - 4)} = 0$

Subtract $6 b$ from $3 b$ .

$\frac{- 3 b - 12 - 8}{(3 b + 4) (b - 4)} = 0$

Subtract $8$ from $- 12$ .

$\frac{- 3 b - 20}{(3 b + 4) (b - 4)} = 0$

Factor $- 1$ out of $- 3 b$ .

$\frac{- (3 b) - 20}{(3 b + 4) (b - 4)} = 0$

Rewrite $- 20$ as $- 1 (20)$ .

$\frac{- (3 b) - 1 (20)}{(3 b + 4) (b - 4)} = 0$

Factor $- 1$ out of $- (3 b) - 1 (20)$ .

$\frac{- (3 b + 20)}{(3 b + 4) (b - 4)} = 0$

Rewrite $- (3 b + 20)$ as $- 1 (3 b + 20)$ .

$\frac{- 1 (3 b + 20)}{(3 b + 4) (b - 4)} = 0$

Move the negative in front of the fraction.

$- \frac{3 b + 20}{(3 b + 4) (b - 4)} = 0$

Step 3

Set the numerator equal to zero.

$3 b + 20 = 0$

Step 4

Solve the equation for $b$ .

Subtract $20$ from both sides of the equation.

$3 b = - 20$

Divide each term in $3 b = - 20$ by $3$ and simplify.

Divide each term in $3 b = - 20$ by $3$ .

$\frac{3 b}{3} = \frac{- 20}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 b}{3} = \frac{- 20}{3}$

Divide $b$ by $1$ .

$b = \frac{- 20}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$b = - \frac{20}{3}$

Solve by Factoring 3/(3b+4)=2/(b-4)

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