Thread: Variable, reciprocals, and fractions..

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.

2 times -3 times (x-2)/2 is -3x + 6 not -3x -6

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.

Originally Posted by rgep

2 times -3 times (x-2)/2 is -3x + 6 not -3x -6

Righto, thanks.