Solve for b b/6+(b+2)/8>=-5/8

$\frac{b}{6} + \frac{b + 2}{8} \geq - \frac{5}{8}$

Step 1

Simplify $\frac{b}{6} + \frac{b + 2}{8}$ .

To write $\frac{b}{6}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{b}{6} \cdot \frac{4}{4} + \frac{b + 2}{8} \geq - \frac{5}{8}$

To write $\frac{b + 2}{8}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{b}{6} \cdot \frac{4}{4} + \frac{b + 2}{8} \cdot \frac{3}{3} \geq - \frac{5}{8}$

Write each expression with a common denominator of $24$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{b}{6}$ by $\frac{4}{4}$ .

$\frac{b \cdot 4}{6 \cdot 4} + \frac{b + 2}{8} \cdot \frac{3}{3} \geq - \frac{5}{8}$

Multiply $6$ by $4$ .

$\frac{b \cdot 4}{24} + \frac{b + 2}{8} \cdot \frac{3}{3} \geq - \frac{5}{8}$

Multiply $\frac{b + 2}{8}$ by $\frac{3}{3}$ .

$\frac{b \cdot 4}{24} + \frac{(b + 2) \cdot 3}{8 \cdot 3} \geq - \frac{5}{8}$

Multiply $8$ by $3$ .

$\frac{b \cdot 4}{24} + \frac{(b + 2) \cdot 3}{24} \geq - \frac{5}{8}$

Combine the numerators over the common denominator.

$\frac{b \cdot 4 + (b + 2) \cdot 3}{24} \geq - \frac{5}{8}$

Simplify the numerator.

Move $4$ to the left of $b$ .

$\frac{4 \cdot b + (b + 2) \cdot 3}{24} \geq - \frac{5}{8}$

Apply the distributive property.

$\frac{4 b + b \cdot 3 + 2 \cdot 3}{24} \geq - \frac{5}{8}$

Move $3$ to the left of $b$ .

$\frac{4 b + 3 \cdot b + 2 \cdot 3}{24} \geq - \frac{5}{8}$

Multiply $2$ by $3$ .

$\frac{4 b + 3 \cdot b + 6}{24} \geq - \frac{5}{8}$

Add $4 b$ and $3 b$ .

$\frac{7 b + 6}{24} \geq - \frac{5}{8}$

Step 2

Multiply both sides by $24$ .

$\frac{7 b + 6}{24} \cdot 24 \geq - \frac{5}{8} \cdot 24$

Step 3

Simplify.

Simplify the left side.

Cancel the common factor of $24$ .

Cancel the common factor.

$\frac{7 b + 6}{24} \cdot 24 \geq - \frac{5}{8} \cdot 24$

Rewrite the expression.

$7 b + 6 \geq - \frac{5}{8} \cdot 24$

Simplify the right side.

Simplify $- \frac{5}{8} \cdot 24$ .

Cancel the common factor of $8$ .

Move the leading negative in $- \frac{5}{8}$ into the numerator.

$7 b + 6 \geq \frac{- 5}{8} \cdot 24$

Factor $8$ out of $24$ .

$7 b + 6 \geq \frac{- 5}{8} \cdot (8 (3))$

Cancel the common factor.

$7 b + 6 \geq \frac{- 5}{8} \cdot (8 \cdot 3)$

Rewrite the expression.

$7 b + 6 \geq - 5 \cdot 3$

Multiply $- 5$ by $3$ .

$7 b + 6 \geq - 15$

Step 4

Solve for $b$ .

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

Subtract $6$ from both sides of the inequality.

$7 b \geq - 15 - 6$

Subtract $6$ from $- 15$ .

$7 b \geq - 21$

Divide each term in $7 b \geq - 21$ by $7$ and simplify.

Divide each term in $7 b \geq - 21$ by $7$ .

$\frac{7 b}{7} \geq \frac{- 21}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 b}{7} \geq \frac{- 21}{7}$

Divide $b$ by $1$ .

$b \geq \frac{- 21}{7}$

Simplify the right side.

Divide $- 21$ by $7$ .

$b \geq - 3$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$b \geq - 3$

Interval Notation:

$[- 3, \infty)$

Solve for b b/6+(b+2)/8>=-5/8

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