Foundations of Math I Drill Booklet - Equations Drill #2, Q3

[SarahM804]

I understand the answer. I don't understand why I cannot solve the problem this way instead:

(10+30+50)/3 = 5 + (x+20+40)/3

First i solve for the left

90/3, which becomes 30. left side is 30. Right side can be broken down a little. Right side is (x + 60)/3

30 = 5 + (x+60)/3

I bring the 5 to the left and subtract it from the 30

25 = (x+60)/3

To get rid of the division of 3, I multiply both sides by 3
Left side: 25*3 = 75
Right side: 3x + 180

75 = 3x + 180
75 - 180 = 3x
-105 = 3x

Divide both sides by 3
x = (-105/3)

Why is this wrong? What rule am I forgetting? Why can't I just solve for the left side, which is really just an integer? Then, bring the 5 over, which is also just an integer?

Thanks!!!

[RonPurewal]

hi, welcome to the forum (i noticed that this is your first post).

but... please take the time to familiarize yourself with the forum rules (= first post in every folder) BEFORE posting. this post violates 2 forum rules:

• it's in the wrong folder (it needs to be posted in the "MPrep non-CAT math" folder);

• you MUST post the ENTIRE TEXT OF THE PROBLEM, exactly as originally provided in the source material.
this is crucial, because we (forum moderators) don't normally have the books on hand while moderating the forum. if you don't post the entire problem, we can't help you. (also, the value of the thread as a future resource is basically eliminated if the problem isn't posted.)

please re-post the COMPLETE PROBLEM -- with the official answer, and with any relevant portions of the answer key -- in the correct folder, along with your questions. thank you.