# Strategies for Solving Linear Equations (are step 1 and 2 mixed up?)

• October 23rd 2010, 12:41 PM
piercedgeek
Strategies for Solving Linear Equations (are step 1 and 2 mixed up?)
According to my book, and some sources I've found on the internet, the common strategy for solving a linear equation is:
1) If fractions are present, multiply each side by the LCD to eliminate them. If decimals are present, multiply each side by a power of 10 to eliminate them.
2) Use the distributive property to remove parentheses.
3) Combine any like terms.
4) Use the addition property of equality to get all variables on one side and numbers on the other.
5) Use the multiplication property of equality to get a single variable on one side.
6) Check by replacing the variable in the original equation with your solution.

My problem is 1) and 2). I've learned the hard way that if you multiply each side by the LCD before distributing to any fractions within parentheses, you get the wrong answer.

Attached is my wrong answer with work shown. To come up with the right answer, at step 1 I needed to distribute the 1/2 and 1/3 to the parenthesis first.
Is there any reason to not do step 2 in the strategy as a general rule before step 1? Why would they be in that order if it sometimes does not work?

Thank you,
Chris

Attachment 19429
• October 23rd 2010, 12:51 PM
Unknown008
Your problem is when distributing the LCD.

$6(2 + 4(y+2) \neq 12 + 8(6y + 12)$

$6(2 + 4(y+2) = 12 + 8(y + 2)$

It's like

6(3x) = 18x

You don't put (18)(6x) = 72x
• October 23rd 2010, 01:05 PM
piercedgeek
So don't distribute the LCD (or factor of 10 in the case of decimals) to inside the parentheses. But would it make a difference if I just started removing all parentheses as step 1 prior to dealing with fractions/decimals with LCD/factor of 10?
• October 23rd 2010, 01:09 PM
Unknown008
If you removed all the parenthesis before distributing the LCD, then you multiply each term, yes.
• October 23rd 2010, 01:24 PM
piercedgeek
Thank you!