Re: Same types of probs as Jalomy


[ Follow Ups ] [ Post Followup ] [ Main Message Board ] [ FAQ ]

Posted by Joel on September 22, 2002 at 20:21:29:

In Reply to: Same types of probs as Jalomy posted by Faith Richards on September 22, 2002 at 19:56:46:

: I have the same types of probs as Jalomy can any one help me solve for x

: (3*x+25/(x+7))-5= 3/x

:
: (2*x-1/x)+1= 4/(x+1)

:
: 4/x - 3/(x+1)=7

Let's take the first one as an example:

(3*x+25/(x+7))-5= 3/x

To clean it up, multiply both sides by the lowest common denominator, which in this case is x*(x+7):
x*(x+7)*[(3*x+25/(x+7))-5] = (3/x)*x*(x+7)

Now multiply that out and CAREFULLY combine like terms & you should end up with something much more manageable. Give it a try.

The same technique works for the other ones.



Follow Ups:



Post a Followup

Name:
E-Mail:

Subject:

Comments:

Optional Link URL:
Link Title:
Optional Image URL:


[ Follow Ups ] [ Post Followup ] [ Main Message Board ] [ FAQ ]