 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   