Question

solve for u.
\\(\frac{6}{7u} = \frac{1}{14} + \frac{7}{2u}\\)
if there is more than one solution, separate them with commas.
if there is no solution, click on
o solution\.
\\(u = \square\\)

Explanation:

Step1: Find common denominator

The denominators are \(7u\), \(14\), and \(2u\). The least common denominator (LCD) of \(7u\), \(14\), and \(2u\) is \(14u\). Multiply each term by \(14u\) to eliminate the denominators:

$$ 14u\times\frac{6}{7u}=14u\times\frac{1}{14}+14u\times\frac{7}{2u} $$

Step2: Simplify each term

Simplify each term after multiplying by \(14u\):

• For the left - hand side: \(14u\times\frac{6}{7u}=\frac{14u\times6}{7u}=2\times6 = 12\)
• For the first term on the right - hand side: \(14u\times\frac{1}{14}=u\)
• For the second term on the right - hand side: \(14u\times\frac{7}{2u}=\frac{14u\times7}{2u}=7\times7 = 49\)

So the equation becomes \(12=u + 49\).

Step3: Solve for \(u\)

Subtract \(49\) from both sides of the equation \(12=u + 49\):
\(u=12 - 49=- 37\)
We need to check if this solution makes the original denominators zero. For the original equation \(\frac{6}{7u}=\frac{1}{14}+\frac{7}{2u}\), when \(u =- 37\), \(7u=7\times(-37)
eq0\) and \(2u=2\times(-37)
eq0\). So \(u=-37\) is a valid solution.

Answer:

\(u=-37\)