Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

How do you solve #5-3/(2x-1)=(x-4)/(2x-1)# and check for extraneous solutions?

1 Answer

Shwetank Mauria

Aug 26, 2016

#x=4/9#

Explanation:

In #5-3/(2x-1)=(x-4)/(2x-1)# let us multiply each term by #(2x-1)#, assuming #2x-1!=0# or #x!=1/2#. We get

#5(2x-1)-3=x-4# or

#10x-5-3=x-4# or

#10x-x=5+3-4# or

#9x=4# or

#x=4/9#

Related questions

See all questions in Extraneous Solutions

Impact of this question

1512 views around the world

You can reuse this answer
Creative Commons License