# solving for x.

• Sep 11th 2009, 05:43 PM
dtownangel
solving for x.
Solve for x
2x/(x-3)=4+(6/(x-3)
2x=4+6
2x=10
x=5

But if I put 5 in for x in the equation it doesnt work.

Where am I going wrong?
• Sep 11th 2009, 05:59 PM
Wilmer
You cannot cancel the (x-3)'s as you did.

Take this case which is similar:
20/4 = 2 + 12/4
You can plainly see that 20 is not equal 2 + 12, right?
• Sep 11th 2009, 06:09 PM
dtownangel
yes
• Sep 11th 2009, 06:16 PM
11rdc11
You are getting confused there is no solution for this problem.
• Sep 11th 2009, 06:18 PM
Wilmer
20/4 = 2 + 12/4

If we change the 2 to 8/4, then:
20/4 = 8/4 + 12/4
20 = 8 + 12
Clear?

Work your problem the same way.
• Sep 11th 2009, 06:37 PM
dtownangel
alright! thanks.
• Sep 11th 2009, 06:57 PM
Wilmer
Quote:

Originally Posted by 11rdc11
You are getting confused there is no solution for this problem.

True; but I think his equation: 2x/(x-3)=4+(6/(x-3)
should be:
2x/(x-3) = 4 + 6x/(x-3)
• Sep 11th 2009, 06:59 PM
dtownangel
well its 6 over x-3
• Sep 11th 2009, 07:03 PM
11rdc11
Quote:

Originally Posted by dtownangel
well its 6 over x-3

Like this?

$\displaystyle \frac{2x}{x-3} = 4 + \frac{6}{x-3}$
• Sep 11th 2009, 07:14 PM
dtownangel
yeah

does x=-3
• Sep 11th 2009, 07:17 PM
11rdc11
Quote:

Originally Posted by dtownangel
yeah

K then the problem has no solution. When you work it out you end up with x= 3 which can't be because you would be dividing by zero which is not allowed
• Sep 11th 2009, 07:20 PM
Wilmer
No; you should get x = 3.
Which means "no solution", since x - 3 = 0; division by 0 is illegal.