#### viva_la_isabel

##### New member

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- Sep 10, 2006

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- 3

i am stuck on this problem:

1/3x + 4 + 2x + 2/3 = 2 2/3x + 1

solve for x

- Thread starter viva_la_isabel
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- Joined
- Sep 10, 2006

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i am stuck on this problem:

1/3x + 4 + 2x + 2/3 = 2 2/3x + 1

solve for x

- Joined
- Feb 4, 2004

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- 15,945

Most of the tutors here tend to be more interested in helping you learn how to do these for yourself. So please help us help you: Post what you've tried so far, even if you think it is wrong.viva_la_isabel said:Please show me the step by step solution

What you have posted could mean the following:viva_la_isabel said:1/3x + 4 + 2x + 2/3 = 2 2/3x + 1

. . . . .\(\displaystyle \L \frac{1}{3x}\, +\, 4\, +\, 2x\, +\, \frac{2}{3}\, =\, \frac{22}{3x}\, +\, 1\)

Is this what you meant? What have you tried? How far have you gotten?

Thank you.

Eliz.

- Joined
- Sep 10, 2006

- Messages
- 3

im stuck. wat do i do next?

1/3 x + 4 + 2x + 2/3 = 2 2/3 x + 1 combine like terms

2 1/3 x + 4 2/3 = 2 2/3 + 1

i know im suppose to solve for x but which x do i solve for. how do i eleminate one.

Were you absent from class when this topic was covered?viva_la_isabel said:im stuck. wat do i do next?

1/3 x + 4 + 2x + 2/3 = 2 2/3 x + 1 combine like terms

2 1/3 x + 4 2/3 = 2 2/3 + 1

i know im suppose to solve for x but which x do i solve for. how do i eleminate one.

Seems so. You need classroom help.