Solve the Inequality for b b/8+(b+3)/6>=-3

$\frac{b}{8} + \frac{b + 3}{6} \geq - 3$

Simplify $\frac{b}{8} + \frac{b + 3}{6}$ .

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

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

To write $\frac{b + 3}{6}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

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

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

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

Multiply $8$ by $3$ .

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

Multiply $\frac{b + 3}{6}$ by $\frac{4}{4}$ .

$\frac{b \cdot 3}{24} + \frac{(b + 3) \cdot 4}{6 \cdot 4} \geq - 3$

Multiply $6$ by $4$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

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

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

Apply the distributive property.

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

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

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

Multiply $3$ by $4$ .

$\frac{3 b + 4 \cdot b + 12}{24} \geq - 3$

Add $3 b$ and $4 b$ .

$\frac{7 b + 12}{24} \geq - 3$

Multiply both sides by $24$ .

$\frac{7 b + 12}{24} \cdot 24 \geq - 3 \cdot 24$

Simplify.

Simplify the left side.

Cancel the common factor of $24$ .

Cancel the common factor.

$\frac{7 b + 12}{24} \cdot 24 \geq - 3 \cdot 24$

Rewrite the expression.

$7 b + 12 \geq - 3 \cdot 24$

Simplify the right side.

Multiply $- 3$ by $24$ .

$7 b + 12 \geq - 72$

Solve for $b$ .

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

Subtract $12$ from both sides of the inequality.

$7 b \geq - 72 - 12$

Subtract $12$ from $- 72$ .

$7 b \geq - 84$

Divide each term in $7 b \geq - 84$ by $7$ and simplify.

Divide each term in $7 b \geq - 84$ by $7$ .

$\frac{7 b}{7} \geq \frac{- 84}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 b}{7} \geq \frac{- 84}{7}$

Divide $b$ by $1$ .

$b \geq \frac{- 84}{7}$

Simplify the right side.

Divide $- 84$ by $7$ .

$b \geq - 12$

The result can be shown in multiple forms.

Inequality Form:

$b \geq - 12$

Interval Notation:

$[- 12, \infty)$

Solve the Inequality for b b/8+(b+3)/6>=-3

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