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

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

Move $\frac{10}{5}$ to the left side of the equation by subtracting it from both sides.

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

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

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Simplify each term.

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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 $\frac{b + 5}{b + 5}$ .

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

Simplify terms.

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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.

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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 $7$ .

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

Simplify with factoring out.

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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 $- (2 b) - 1 (3)$ .

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

Simplify the expression.

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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.

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

Simplify each term.

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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 $\frac{b + 5}{b + 5}$ .

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

Simplify terms.

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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.

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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 $7$ .

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

Simplify with factoring out.

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

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

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

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

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

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

Simplify the expression.

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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.

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

Find the LCD of the terms in the equation.

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Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$b + 5, 1$

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of $1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

The factor for $b + 5$ is $b + 5$ itself.

$(b + 5) = b + 5$

$(b + 5)$ occurs $1$ time.

The LCM of $b + 5$ is the result of multiplying all factors the greatest number of times they occur in either term.

$b + 5$

Multiply each term by $b + 5$ and simplify.

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Multiply each term in $- \frac{2 b + 3}{b + 5} = 0$ by $b + 5$ in order to remove all the denominators from the equation.

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

Simplify $- \frac{2 b + 3}{b + 5} \cdot (b + 5)$ .

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Cancel the common factor of $b + 5$ .

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Move the leading negative in $- \frac{2 b + 3}{b + 5}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

$- (2 b + 3) = 0 \cdot (b + 5)$

Apply the distributive property.

$- (2 b) - 1 \cdot 3 = 0 \cdot (b + 5)$

Multiply.

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Multiply $2$ by $- 1$ .

$- 2 b - 1 \cdot 3 = 0 \cdot (b + 5)$

Multiply $- 1$ by $3$ .

$- 2 b - 3 = 0 \cdot (b + 5)$

Multiply $0$ by $b + 5$ .

$- 2 b - 3 = 0$

Solve the equation.

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Add $3$ to both sides of the equation.

$- 2 b = 3$

Divide each term by $- 2$ and simplify.

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

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

Cancel the common factor of $- 2$ .

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Cancel the common factor.

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

Divide $b$ by $1$ .

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

Move the negative in front of the fraction.

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

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

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