2 times -3 times (x-2)/2 is -3x + 6 not -3x -6
I came across another question which has just stumped me because I worked it over about 5 times and I couldn't figure out how the book got the answer. I've obtained messed up answers and none have them have been...
Answer: 6
Problem:
2x - 3(x-2)/2 = 7 - x-3/3
The thing that probably messed me up the most was the fact that I didn't know how to get the variable fractions out of the way.
I really don't remember the book covering that, therefore I sucked at it.
I did somehow try to solve it:
2(2x - 3(x-2)/2) = (7 - x-3/3)2
4x - 3x - 6 = 14 - 2x-6/3
x - 6 = 14... +6+14...
x = 20 - 2x-6/6
3(x) = (20 - 2x-6/3)3
3x = 60 - 2x - 6
+2x
8x = 54
8/8 = 54/8
= 6.75
So basically i'm screwing up somewhere and I don't know where but I'm not getting the right answer.
Here is one way
My favorite way in dealing with fractions in equations is to eliminate the fractions first. Make all numbers as whole numbers.
In doing that, I multiply both sides of the equations by a common multiplier. This common multiplier is the product of all the denominators, but no denominator is to repeat or appear more than once.
2x - 3(x-2)/2 = 7 - x-3/3
Rewriting that,
2x - 3(x-2)/2 = 7 -(x-3)/3
There are two fractions, so two denominators: 2 and 3.
Clear the fractions, multiply both sides by 2*3,
2*3*2x -3*3(x-2) = 2*3*7 -2*(x-3)
12x -9(x-2) = 42 -2(x-3)
12x -9x +18 = 42 -2x +6
Collect like terms, or isolate the x,
12x -9x +2x = 42 +6 -18
5x = 30
x = 30/5 = 6 ----answer.