Algebraic expression #10

[mathwannabe]

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...

 

[mathwannabe]

What do the colons mean?

Colons are division signs. We use \(\displaystyle :\) to signify division a lot more than we use \(\displaystyle /\).

 

[mathwannabe]

I am not sure I understand the problem

\(\displaystyle Let\ w = \dfrac{\frac{a}{a - b}+\frac{b}{a}}{a - (a - \frac{b}{a+1})} =\)

\(\displaystyle \dfrac{\frac{a^2 + ab - b^2}{a(a - b)}}{\frac{b}{a + 1}} = \dfrac{(a + 1)(a^2 + ab - b^2)}{ab(a - b)}.\) You get (b -a) in the denominator.

\(\displaystyle Let\ x = \dfrac{\frac{a}{b}+\frac{a(1 + b) - b^3}{b(ab -1)}}{1 -\frac{1}{1 - ab}} =\)

\(\displaystyle \dfrac{\frac{a^2b^2 - ab + ab + ab^2 - b^4}{b^2(ab - 1)}}{\frac{-ab}{1 -ab}} =\) Why expand with \(\displaystyle b^2(ab-1)\)? Isn't LCD just \(\displaystyle b(ab-1)\)? So we expand the left addend with \(\displaystyle (ab-1)\) instead with \(\displaystyle b(ab-1)\) and leave the right addend intact instead of expanding it with \(\displaystyle b\).

\(\displaystyle \dfrac{b^2(a^2 + a - b^2)}{b^2(ab -1)} * \dfrac{ab - 1}{ab} =\)

\(\displaystyle \dfrac{a^2 + a - b^2}{ab}.\)

Problem: Reduce \(\displaystyle \dfrac{\frac{w}{x}}{\frac{a + 1}{a - b}} = w * \dfrac{1}{x} * \dfrac{a - b}{a + 1} =\)

\(\displaystyle \dfrac{(a + 1)(a^2 + ab - b^2)}{ab(a - b)} * \dfrac{ab}{a^2 + a - b^2} * \dfrac{a - b}{a + 1} = 1.\)

I get \(\displaystyle ab(b-a)\) instead of \(\displaystyle ab(a-b)\). I'm gonna sit down and do it again.

 

[mathwannabe]

Because I am too old and so too lazy to look for LCDs; I just multiply denominators together.

It makes no difference of course; what is important is to have identical denominators. You get to exactly the same place.

Yes, off course, it really makes no difference. It just confused me, I thought it was maybe something I was doing wrong and you knew some super secret kung fu move there

 

[mathwannabe]

Your problem is here, not in the expansion in the previous post.

\(\displaystyle \dfrac{\frac{a}{a - b}+\frac{b}{a}}{a - (a - \frac{b}{a+1})} =\)

\(\displaystyle \dfrac{\frac{a^2 + ab - b^2}{a(a - b)}}{a - a + \frac{b}{a+1}} =\)

\(\displaystyle \dfrac{\frac{a^2 + ab -b^2}{a(a - b)}}{\frac{b}{a + 1}} =\)

\(\displaystyle \dfrac{(a + 1)(a^2 + ab - b^2)}{ab(a - b)}.\)

But now I see an error in my original solution

\(\displaystyle a^2 + ab - b^2\) does not cancel \(\displaystyle a^2 + a - b^2.\)

Now I know exactly what I was doing wrong.

Basicly, I was wrong adding the terms in the denominator:

\(\displaystyle a-(a-\frac{b}{a+1})\)

I first added the terms in the bracket and then forgot to change the sign after I removed the brackets. Instead, I should have just get rid of the parentheses and change signs right away. What's troubling is that I made the same mistake few times in a row. I guess it's true what they say about mistakes having tendency to be repeated.

That's a problem from one of the previous exams. \(\displaystyle -1\) is one of the offered answers, it's just not the correct one. I guess they know what mistakes students make often times

Anyway, thanks Jeff, your help is highly valued, as always.