Rewrite using positive exponents
(2 / 2 + x^-1) - 3

Question

Rewrite using positive exponents
(2 / 2 + x^-1)  - 3

mathstudent55 · Answer

$a^{-n} = \dfrac{1}{a^n} $ for $a \ne 0$.

anonymous · Answer

thank you :) would you help me apply that formula to the problem please

mathstudent55 · Answer

I'm not sure how to read the problem.
Is it

$\dfrac{2}{2 + x^{-1}} - 3$    ?

anonymous · Answer

yes

mathstudent55 · Answer

Then we replace $x^{-1} $ with what you get using the formula above:

$ \dfrac{2}{2 + x^{-1}} - 3$

$ =\dfrac{2}{2 + \frac{1}{x}} - 3 $ 

Now multiply the numerator and denominator of the fraction by x to get rid of the fraction 1/x in the denominator.

anonymous · Answer

ok thank you, so the answer would be (2x / 2x + 1) - 3?

mathstudent55 · Answer

$ =\dfrac{2x}{2x+ 1}- 3 $ 

That is correct. To go further, you need to do the subtraction using a common denominator.

mathstudent55 · Answer

$ =\dfrac{2x}{2x+ 1}- \dfrac{3}{1} \dfrac{2x+1}{2x+1}$ 

$ =\dfrac{2x}{2x+ 1}- \dfrac{6x+3}{2x+1}$ 

$ =\dfrac{2x - (6x+3)}{2x+ 1} $

$ =\dfrac{2x - 6x-3}{2x+ 1} $

$ =\dfrac{-4x-3}{2x+ 1} $

anonymous · Answer

thank you so much for explaining it so clearly! would you mind helping with one more I have pages of them to do...

a^-1 - b^-1 / b^-2 - a^ -2

mathstudent55 · Answer

You mean

$ \dfrac{a^{-1} - b^{-1} } {b^{-2} - a^{-2} } $   ?

anonymous · Answer

yes

anonymous · Answer

would the answer be b^2 - a^2 / a^1 - b^1

mathstudent55 · Answer

I don't know. I haven't solved it yet.
Start by changing each negative exponent to a fraction using the formula above.

mathstudent55 · Answer

$ \dfrac{a^{-1} - b^{-1} } {b^{-2} - a^{-2} }$

$ = \dfrac{ \frac{1}{a}  - \frac{1}{b}   } {\frac{1}{b^2} - \frac{1}{a^2} } $

Now multiply the numerator and denominator by $a^2b^2$ to get rid of the fractions in the numerator and denominator.

anonymous · Answer

hmm ok i got ab^2 - a^2b / a^2 - b^2 ...is that right?

mathstudent55 · Answer

$ = \dfrac{ab^2 - a^2b}{a^2 - b^2} $

Yes, I got the same, but we're not done yet.

Now factor the numerator and denominator.

mathstudent55 · Answer

$ = \dfrac{ab(b - a)}{(a + b)(a - b)} $

It looks like nothing in the numerator and denominator cancels out, but look at the next step:

$ = \dfrac{-ab(a - b)}{(a + b)(a - b)} $

$ = \dfrac{-ab\cancel{(a - b)}}{(a + b)\cancel{(a - b)}} $

$ = \dfrac{-ab}{a + b} $

mathstudent55 · Answer

This is very useful:

a - b = -(b - a)

It's easy to show it's true:

a - b = -(-a + b) = -(b - a)

anonymous · Answer

I would never have thought of that, thanks

mathstudent55 · Answer

wlcm