## Need help verifying an anwer to this problem.

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
rogermiranda
Posts: 19
Joined: Fri Apr 30, 2010 6:01 pm
Contact:

### Need help verifying an anwer to this problem.

Complex Fraction:
a^2 / b^2 - 1 denominator 1/a + 1/b = a^2 - b^2 / b^2 denominator b+a / ab
= a^2 - b^2 / b^2 * ab / b+a = a(a-b) / b.
But my textbook has the answer a(a-b). So what happened to the denominator b?

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
Please reply showing the expression with grouping symbols. For instance, does "a^2 / b^2 - 1" mean "(a2/b2) - 1" or "a2/(b2 - 1)"?

Thank you!

rogermiranda
Posts: 19
Joined: Fri Apr 30, 2010 6:01 pm
Contact:

### Re: Need help verifying an anwer to this problem.

Here is the Complex Fraction:
Numerator= (a^2 / b^2) - 1
Denominator = (1/a + b/a)

So, finding the LCD I have
Numerator= (a^2 - b^2 / b^2)
Denominator = (b + a / ab)

Next I multiply by the reciprocal
(a^2 - b^2 / b^2) * (ab / b + a)
[(a + b)(a - b) / b^2)] * (ab / b + a)

Simplifying, I am left with
(a - b / b) * a

So the answer I get is
Numerator= a(a - b)
Denominator = b but my textbook has the answer a(a - b). So what happened to the denominator b?

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
Here is the Complex Fraction:
Numerator= (a^2 / b^2) - 1
Denominator = (1/a + b/a)
I will assume the following:

. . . . .$\frac{\left(\frac{a^2}{b^2}\, -\, 1\right)}{\left(\frac{1}{a}\, +\,\frac{b}{a}\right)}$

I think you then went the route of working the numerator and denominator separately. But I don't understand what you did in the denominator, since you were given two fractions with the same denominator...? I think the next step should have been as follows:

. . . . .$\frac{\left(\frac{a^2\, -\, b^2}{b^2}\right)}{\left(\frac{1\, +\, b}{a}\right)}$

rogermiranda
Posts: 19
Joined: Fri Apr 30, 2010 6:01 pm
Contact:

### Re: Need help verifying an anwer to this problem.

Yes you have written the problem correctly.
Except the denomitor of the complex fraction is 1 over a + 1 over b (1/a + 1/b). So the LCD is ab.
So the denominator of the complex fraction is now b + a over ab (b + a / ab).

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
... the denomitor of the complex fraction is 1 over a + 1 over b (1/a + 1/b).
Ah; so the expression is as follows...?

. . . . .$\frac{\left(\frac{a^2}{b^2}\, -\, 1\right)}{\left(\frac{1}{a}\, +\,\frac{1}{b}\right)}$

Then the LCD's would give:

. . . . .$\frac{\left(\frac{a^2\, -\, b^2}{b^2}\right)}{\left(\frac{b\, +\, a}{ab}\right)}$

Reverse the one addition to get the standard order:

. . . . .$\frac{\left(\frac{a^2\, -\, b^2}{b^2}\right)}{\left(\frac{a\, +\, b}{ab}\right)}$

Then flip-n-multiply:

. . . . .$\left(\frac{a^2\, -\, b^2}{b^2}\right)\, \times\, \left(\frac{ab}{a\, +\, b}\right)$

Factor the one numerator, cancel the common factors, and simplify to get the final answer.

My answer matches yours, by the way, not the book's. At a guess, the solution was typoed.