Question

How do you solve and check for extraneous solutions in `4/v + 1/5 = 1`?

Answer 1

Isolate the term containing the variable; multiply to clear the denominators and divide by the constant to get `v=5` then plug the solution back into the original equation to verify.

Explanation:

Given `4/v+1/5=1`

Subtract `1/5` from both sides
`color(white)("XXXX")` `4/v = 4/5`
(At this point it should be obvious that `v=5`, but continuing on...)
Multiply both sides by `5v`
`color(white)("XXXX")` `4*5 = 4v`
Divide by `5`
`color(white)("XXXX")` `v=5`

Verify that this is not an extraneous solution by substituting `color(red)(5)` for `v` in the original equation:

`color(white)("XXXX")` `4/color(red)(5) + 1/5 = 1`