Solve by Factoring 7/(b+5)=10/5

$\frac{7}{b + 5} = \frac{10}{5}$

Step 1

Subtract $\frac{10}{5}$ from both sides of the equation.

$\frac{7}{b + 5} - \frac{10}{5} = 0$

Step 2

Simplify $\frac{7}{b + 5} - \frac{10}{5}$ .

Simplify each term.

Divide $10$ by $5$ .

$\frac{7}{b + 5} - 1 \cdot 2 = 0$

Multiply $- 1$ by $2$ .

$\frac{7}{b + 5} - 2 = 0$

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{b + 5}{b + 5}$ .

$\frac{7}{b + 5} - 2 \cdot \frac{b + 5}{b + 5} = 0$

Combine $- 2$ and $\frac{b + 5}{b + 5}$ .

$\frac{7}{b + 5} + \frac{- 2 (b + 5)}{b + 5} = 0$

Combine the numerators over the common denominator.

$\frac{7 - 2 (b + 5)}{b + 5} = 0$

Simplify the numerator.

Apply the distributive property.

$\frac{7 - 2 b - 2 \cdot 5}{b + 5} = 0$

Multiply $- 2$ by $5$ .

$\frac{7 - 2 b - 10}{b + 5} = 0$

Subtract $10$ from $7$ .

$\frac{- 2 b - 3}{b + 5} = 0$

Factor $- 1$ out of $- 2 b$ .

$\frac{- (2 b) - 3}{b + 5} = 0$

Rewrite $- 3$ as $- 1 (3)$ .

$\frac{- (2 b) - 1 (3)}{b + 5} = 0$

Factor $- 1$ out of $- (2 b) - 1 (3)$ .

$\frac{- (2 b + 3)}{b + 5} = 0$

Rewrite $- (2 b + 3)$ as $- 1 (2 b + 3)$ .

$\frac{- 1 (2 b + 3)}{b + 5} = 0$

Move the negative in front of the fraction.

$- \frac{2 b + 3}{b + 5} = 0$

Step 3

Set the numerator equal to zero.

$2 b + 3 = 0$

Step 4

Solve the equation for $b$ .

Subtract $3$ from both sides of the equation.

$2 b = - 3$

Divide each term in $2 b = - 3$ by $2$ and simplify.

Divide each term in $2 b = - 3$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $b$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Solve by Factoring 7/(b+5)=10/5

Related Questions

Support LCMGCF.com by disabling your adblocker.