SOLUTION: Please help me answer this question: {{{(1/8)-2/3}}} I started this: (1/8)-2/3 = 1/(1/8)-2/3) = 1 x (8/1)-2/3 this is where I get stuck. Thanks.

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: Please help me answer this question: {{{(1/8)-2/3}}} I started this: (1/8)-2/3 = 1/(1/8)-2/3) = 1 x (8/1)-2/3 this is where I get stuck. Thanks.      Log On


   



Question 121726: Please help me answer this question:
%281%2F8%29-2%2F3
I started this:
(1/8)-2/3 = 1/(1/8)-2/3) = 1 x (8/1)-2/3 this is where I get stuck.
Thanks.

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Your problem is either subtracting fractions or taking fractions to a fractional power.
We can look at both.
1%2F8-2%2F3
Subtractng fractions.
Whenever you add or subtract fractions you need to have the same denominator.
The easiest way to get a common denominator is to make it the product of your two denominators.
D=8%2A3=24
Now find each fraction as an equivalent fraction with denominator of 24.
1%2F8=x%2F24
24%2A%281%2F8%29=x
x=3
1%2F8=3%2F24
and
2%2F3=y%2F24
24%2A%282%2F3%29=y
y=16
2%2F3=16%2F24
Now you can subtract.
1%2F8-2%2F3=3%2F24-16%2F24
1%2F8-2%2F3=%283-16%29%2F24
1%2F8-2%2F3=-%2813%2F24%29
Taking fractions to a fractional power.
%281%2F8%29%5E%28-2%2F3%29
You can also look at 1/8 as 8%5E%28-1%29.
%281%2F8%29%5E%28-2%2F3%29=%288%5E%28-1%29%29%5E%28-2%2F3%29
From your exponentiation rules,
%28x%5EM%29%5EN=x%5E%28M%2AN%29
Then
%281%2F8%29%5E%28-2%2F3%29=%288%5E%28-1%29%29%5E%28-2%2F3%29
%281%2F8%29%5E%28-2%2F3%29=%288%29%5E%282%2F3%29
%281%2F8%29%5E%28-2%2F3%29=%282%29%5E%282%29
%281%2F8%29%5E%28-2%2F3%29=4