Question

solve for (u).

6 + \frac{1}{u - 1} = -\frac{6}{u - 2}

if there is more than one solution, separate them with commas.
if there is no solution, click on
o solution\.

Explanation:

Identify domain restrictions

To ensure the rational expressions are defined, the denominators cannot be zero.

$$ u - 1 eq 0 \implies u eq 1 $$

$$ u - 2 eq 0 \implies u eq 2 $$

Clear the denominators

Using the Least Common Multiple knowledge point

$$ \text{LCM} = (u - 1)(u - 2) $$

$$ 6(u - 1)(u - 2) + 1(u - 2) = -6(u - 1) $$

Expand and simplify the equation

Using the Factoring Polynomials knowledge point

$$ 6(u^2 - 3u + 2) + u - 2 = -6u + 6 $$

$$ 6u^2 - 18u + 12 + u - 2 = -6u + 6 $$

$$ 6u^2 - 17u + 10 = -6u + 6 $$

$$ 6u^2 - 11u + 4 = 0 $$

Solve the quadratic equation

We factor the quadratic expression:

$$ 6u^2 - 8u - 3u + 4 = 0 $$

$$ 2u(3u - 4) - 1(3u - 4) = 0 $$

$$ (2u - 1)(3u - 4) = 0 $$

$$ u = \frac{1}{2} \quad \text{or} \quad u = \frac{4}{3} $$

Check for extraneous solutions

We compare our solutions with the domain restrictions \(u
eq 1\) and \(u
eq 2\).
Both \(u = \frac{1}{2}\) and \(u = \frac{4}{3}\) are valid solutions.

Answer:

Solve for \(u\).

$$6 + \frac{1}{u - 1} = -\frac{6}{u - 2}$$

If there is more than one solution, separate them with commas.
If there is no solution, click on "No solution".

\(u =\) <blank>\(\frac{1}{2}, \frac{4}{3}\)</blank>