What is the answer to (u+2)/(3u-1)+1=(6-u)/(u+1) ?

The answer to (u+2)/(3u-1)+1=(6-u)/(u+1) is u=1

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(u+2)/(3u-1)+1=(6-u)/(u+1)

\frac{u + 2}{3 u - 1} + 1 = \frac{6 - u}{u + 1}

u = 1

Solution steps

\frac{u + 2}{3 u - 1} + 1 = \frac{6 - u}{u + 1}

Multiply by LCM

(u + 2) (u + 1) + (u + 1) (3 u - 1) = (6 - u) (3 u - 1)

Solve

(u + 2) (u + 1) + (u + 1) (3 u - 1) = (6 - u) (3 u - 1) : u = 1

u = 1

Verify Solutions

Find undefined (singularity) points:

u = \frac{1}{3}, u = - 1

Combine undefined points with solutions:

u = 1

\frac{1}{x ^{2}} + \frac{3}{x} - 10 = 0

\frac{50}{x + 5} = \frac{x}{10} + 4.5

\frac{2}{x} + \frac{x + 6}{4} = 0

\frac{10000}{x} = 9

\frac{4}{x} + \frac{5}{x + 3} = 3

What is the answer to (u+2)/(3u-1)+1=(6-u)/(u+1) ?
The answer to (u+2)/(3u-1)+1=(6-u)/(u+1) is u=1
Solve for u: (u+2)/(3u-1)+1=(6-u)/(u+1)
The solution is u=1