Simplify (3+v/2)/(v/4-9/v)

$\frac{3 + \frac{v}{2}}{\frac{v}{4} - \frac{9}{v}}$

Multiply the numerator and denominator of the complex fraction by $4 v$ .

Multiply $\frac{3 + \frac{v}{2}}{\frac{v}{4} - \frac{9}{v}}$ by $\frac{4 v}{4 v}$ .

$\frac{4 v}{4 v} \cdot \frac{3 + \frac{v}{2}}{\frac{v}{4} - \frac{9}{v}}$

Combine.

$\frac{4 v (3 + \frac{v}{2})}{4 v (\frac{v}{4} - \frac{9}{v})}$

Apply the distributive property.

$\frac{4 v \cdot 3 + 4 v \frac{v}{2}}{4 v \frac{v}{4} + 4 v (- \frac{9}{v})}$

Simplify by cancelling.

Cancel the common factor of $2$ .

Factor $2$ out of $4 v$ .

$\frac{4 v \cdot 3 + 2 (2 v) \frac{v}{2}}{4 v \frac{v}{4} + 4 v (- \frac{9}{v})}$

Cancel the common factor.

$\frac{4 v \cdot 3 + 2 (2 v) \frac{v}{2}}{4 v \frac{v}{4} + 4 v (- \frac{9}{v})}$

Rewrite the expression.

$\frac{4 v \cdot 3 + 2 v \cdot v}{4 v \frac{v}{4} + 4 v (- \frac{9}{v})}$

Raise $v$ to the power of $1$ .

$\frac{4 v \cdot 3 + 2 (v^{1} v)}{4 v \frac{v}{4} + 4 v (- \frac{9}{v})}$

Raise $v$ to the power of $1$ .

$\frac{4 v \cdot 3 + 2 (v^{1} v^{1})}{4 v \frac{v}{4} + 4 v (- \frac{9}{v})}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{4 v \cdot 3 + 2 v^{1 + 1}}{4 v \frac{v}{4} + 4 v (- \frac{9}{v})}$

Add $1$ and $1$ .

$\frac{4 v \cdot 3 + 2 v^{2}}{4 v \frac{v}{4} + 4 v (- \frac{9}{v})}$

Cancel the common factor of $4$ .

Factor $4$ out of $4 v$ .

$\frac{4 v \cdot 3 + 2 v^{2}}{4 (v) \frac{v}{4} + 4 v (- \frac{9}{v})}$

Cancel the common factor.

$\frac{4 v \cdot 3 + 2 v^{2}}{4 v \frac{v}{4} + 4 v (- \frac{9}{v})}$

Rewrite the expression.

$\frac{4 v \cdot 3 + 2 v^{2}}{v \cdot v + 4 v (- \frac{9}{v})}$

Raise $v$ to the power of $1$ .

$\frac{4 v \cdot 3 + 2 v^{2}}{v^{1} v + 4 v (- \frac{9}{v})}$

Raise $v$ to the power of $1$ .

$\frac{4 v \cdot 3 + 2 v^{2}}{v^{1} v^{1} + 4 v (- \frac{9}{v})}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{4 v \cdot 3 + 2 v^{2}}{v^{1 + 1} + 4 v (- \frac{9}{v})}$

Add $1$ and $1$ .

$\frac{4 v \cdot 3 + 2 v^{2}}{v^{2} + 4 v (- \frac{9}{v})}$

Cancel the common factor of $v$ .

Move the leading negative in $- \frac{9}{v}$ into the numerator.

$\frac{4 v \cdot 3 + 2 v^{2}}{v^{2} + 4 v \frac{- 9}{v}}$

Factor $v$ out of $4 v$ .

$\frac{4 v \cdot 3 + 2 v^{2}}{v^{2} + v \cdot 4 \frac{- 9}{v}}$

Cancel the common factor.

$\frac{4 v \cdot 3 + 2 v^{2}}{v^{2} + v \cdot 4 \frac{- 9}{v}}$

Rewrite the expression.

$\frac{4 v \cdot 3 + 2 v^{2}}{v^{2} + 4 \cdot - 9}$

Multiply $4$ by $- 9$ .

$\frac{4 v \cdot 3 + 2 v^{2}}{v^{2} - 36}$

Simplify the numerator.

Factor $2 v$ out of $4 v \cdot 3 + 2 v^{2}$ .

Factor $2 v$ out of $4 v \cdot 3$ .

$\frac{2 v (2 \cdot 3) + 2 v^{2}}{v^{2} - 36}$

Factor $2 v$ out of $2 v^{2}$ .

$\frac{2 v (2 \cdot 3) + 2 v (v)}{v^{2} - 36}$

Factor $2 v$ out of $2 v (2 \cdot 3) + 2 v (v)$ .

$\frac{2 v (2 \cdot 3 + v)}{v^{2} - 36}$

Multiply $2$ by $3$ .

$\frac{2 v (6 + v)}{v^{2} - 36}$

Simplify the denominator.

Rewrite $36$ as $6^{2}$ .

$\frac{2 v (6 + v)}{v^{2} - 6^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = v$ and $b = 6$ .

$\frac{2 v (6 + v)}{(v + 6) (v - 6)}$

Cancel the common factor of $6 + v$ and $v + 6$ .

Reorder terms.

$\frac{2 v (v + 6)}{(v + 6) (v - 6)}$

Cancel the common factor.

$\frac{2 v (v + 6)}{(v + 6) (v - 6)}$

Rewrite the expression.

$\frac{2 v}{v - 6}$

Simplify (3+v/2)/(v/4-9/v)

Related Questions

Support LCMGCF.com by disabling your adblocker.