Question

solve for v.
\\(\frac{v - 2}{8v} - \frac{1}{12} = \frac{1}{v}\\)
if there is more than one solution, separate them with commas.
if there is no solution, click on
o solution\.
\\(v = \square\\)

Explanation:

Step1: Find common denominator

The denominators are \(8v\), \(12\), and \(v\). The least common denominator (LCD) of \(8v\), \(12\), and \(v\) is \(24v\). Multiply each term by \(24v\) to eliminate the denominators:

$$ 24v\times\frac{v - 2}{8v}-24v\times\frac{1}{12}=24v\times\frac{1}{v} $$

Simplify each term:

• For the first term: \(24v\times\frac{v - 2}{8v}=\frac{24v(v - 2)}{8v}=3(v - 2)\) (the \(v\) cancels out, and \(24\div8 = 3\))
• For the second term: \(24v\times\frac{1}{12}=2v\) (since \(24\div12 = 2\))
• For the third term: \(24v\times\frac{1}{v}=24\) (the \(v\) cancels out)

So the equation becomes:

$$ 3(v - 2)-2v = 24 $$

Step2: Expand and simplify

Expand \(3(v - 2)\):

$$ 3v-6 - 2v=24 $$

Combine like terms (\(3v-2v\)):

$$ v-6 = 24 $$

Step3: Solve for \(v\)

Add \(6\) to both sides of the equation:

$$ v-6 + 6=24 + 6 $$

$$ v = 30 $$

We need to check if this solution makes the original denominators zero. The original denominators are \(8v\), \(12\), and \(v\). When \(v = 30\), \(8v=240
eq0\), \(12
eq0\), and \(v = 30
eq0\). So \(v = 30\) is a valid solution.

Answer:

\(v = 30\)