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

1 Answer
Aug 14, 2015

Answer:

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#