mathwannabe
Junior Member
- Joined
- Feb 20, 2012
- Messages
- 122
Hello Everybody
I have this algebraic expression I need to fix, I have done it 3 times and my result is always \(\displaystyle -1\). The book says it should be \(\displaystyle +1\). I would be extremely grateful if someone could point me to where I am making the mistake.
1) \(\displaystyle (\dfrac{\frac{a}{a-b}+\frac{b}{a}}{a-(a-\frac{b}{a+1})}:\dfrac{\frac{a}{b}+\frac{a(1+b)-b^3}{b(ab-1)}}{1-\frac{1}{1-ab}}):\dfrac{a+1}{a-b}\)
The following step is after I get rid of double fractions:
\(\displaystyle \dfrac{(a+1)(a^2+ab-b^2)}{ab(b-a)}:\dfrac{a^2+ab-b^2}{ab}:\dfrac{a+1}{a-b}\)
I end up with:
\(\displaystyle \dfrac{a-b}{b-a}=-1\)
Knowing myself, I won't be able to sleep if I don't get this right...
I have this algebraic expression I need to fix, I have done it 3 times and my result is always \(\displaystyle -1\). The book says it should be \(\displaystyle +1\). I would be extremely grateful if someone could point me to where I am making the mistake.
1) \(\displaystyle (\dfrac{\frac{a}{a-b}+\frac{b}{a}}{a-(a-\frac{b}{a+1})}:\dfrac{\frac{a}{b}+\frac{a(1+b)-b^3}{b(ab-1)}}{1-\frac{1}{1-ab}}):\dfrac{a+1}{a-b}\)
The following step is after I get rid of double fractions:
\(\displaystyle \dfrac{(a+1)(a^2+ab-b^2)}{ab(b-a)}:\dfrac{a^2+ab-b^2}{ab}:\dfrac{a+1}{a-b}\)
I end up with:
\(\displaystyle \dfrac{a-b}{b-a}=-1\)
Knowing myself, I won't be able to sleep if I don't get this right...