Quandaries and Queries
  

 

Hi! My name is Emma and the level of the question I need help on is middle (I am in 9th grade).
I do not get how to "solve for x". For example:

3x - 5 - (4x - 8) = 2(2 - 3x) + 29

For the above question, I always get confused on what steps to take first, second, third, etc.

I am in regular 9th grade math, and all of my classmates get how to do it and I don't.

Please help!
 

 

Hi Emma,

You first look inside the parentheses to see if you can do any simplification. Inside the parentheses in this problem are

2 - 3 x and 4x - 8

Neither of these can be simplified so the next step is to perform any multiplications. On the right side you have

2(2 - 3x) = 2 2 - 2 3x = 4 - 6x

On the left side -(4x - 8) can be interpreted as (-1)(4x - 8) so

-(4x - 8) = (-1)( 4x - 8) = (-1) (4x) - (-1)(8) = -4x + 8

Thus, so far I have

3x - 5 - (4x - 8) = 2(2 - 3x) + 29
3x - 5 - 4x + 8 = 4 - 6x + 29

Now on each side collect the terms with an x in them and collect the terms without an x to get

3x - 4x -5 + 8 = 4 + 29- 6x
-x + 3 = 33 - 6x

You want to end up with an expression of the form "x = some number" so you have to get all the terms with an x in them on the left of the equation and all the remaining terms on the right. If I add 6x to both sides I will eliminate the -6x term from the right side

-x + 3 = 33 - 6x
-x + 3 + 6x = 33 - 6x + 6x
5x + 3 = 33

To eliminate the 3 from the left side, add -3 to both sides

5x + 3 = 33
5x + 3 - 3 = 33 -3
5x = 30

Finally divide both sides by 5 to get

5x = 30
 5x/5 = 30/5
x = 6

Penny