1. ## Simplify fractions

This problem is driving me nuts... it should be rather easy but...

$\dfrac{\dfrac{1}{a}-\dfrac{1}{a+b}}{\dfrac{1}{b}-\dfrac{1}{a+b}} =$

$\dfrac{\dfrac{\dfrac{b}{a}}{a+b}}{\dfrac{\dfrac{a} {b}}{a+b}}} =$

$\dfrac{\dfrac{b}{a}}{a+b}*\dfrac{a+b}{\dfrac{a}{b} }=$

$\dfrac{\dfrac{b}{a}*(a+b)}{(a+b)*\dfrac{a}{b}}=$

Think the above is correct, but not the following:

$\dfrac{\dfrac{ab+{b}^{2}}{a}}{\dfrac{ab+{a}^{2}}{a }}=$

$\frac{ab+{b}^{2}}{a}*\frac{a}{ab+{a}^{2}}=$

$\frac{{a}^{2}b+a{b}^{2}}{{a}^{2}b+{a}^{3}}=$

$\dfrac{ab+{b}^{2}}{ab+{a}^{2}}$ which is not correct the answer should be $\dfrac{{b}^{2}}{{a}^{2}}$

2. Originally Posted by TriForce
This problem is driving me nuts... it should be rather easy but...

$\dfrac{\dfrac{1}{a}-\dfrac{1}{a+b}}{\dfrac{1}{b}-\dfrac{1}{a+b}} =$

$\dfrac{\dfrac{\dfrac{b}{a}}{a+b}}{\dfrac{\dfrac{a} {b}}{a+b}}} =$

$\dfrac{\dfrac{b}{a}}{a+b}*\dfrac{a+b}{\dfrac{a}{b} }=$

$\dfrac{\dfrac{b}{a}*(a+b)}{(a+b)*\dfrac{a}{b}}=$

Think the above is correct, but not the following:

$\dfrac{\dfrac{ab+{b}^{2}}{a}}{\dfrac{ab+{a}^{2}}{a }}=$

$\frac{ab+{b}^{2}}{a}*\frac{a}{ab+{a}^{2}}=$

$\frac{{a}^{2}b+a{b}^{2}}{{a}^{2}b+{a}^{3}}=$

$\dfrac{ab+{b}^{2}}{ab+{a}^{2}}$ which is not correct the answer should be $\dfrac{{b}^{2}}{{a}^{2}}$
I don't understand what you've done. Here is the method I would have used:

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

Cancel the term in parantheses and you're done.

3. Hello, TriForce!

$\dfrac{\dfrac{1}{a}-\dfrac{1}{a+b}}{\dfrac{1}{b}-\dfrac{1}{a+b}} =$

Multiply numerator and denominator by the LCD, . $ab(a+b)$

$\frac{ab(a+b)\left(\dfrac{1}{a} - \dfrac{1}{a+b}\right)}{ab(a+b)\left(\dfrac{1}{b} - \dfrac{1}{a+b}\right)} \;=\;\dfrac{b(a+b) - ab}{a(a+b) - ab} \;=\;\frac{ab + b^2 - ab}{a^2 + ab - ab} \;=\;\frac{b^2}{a^2}$

4. Originally Posted by earboth
I don't understand what you've done. Here is the method I would have used:

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

Cancel the term in parantheses and you're done.
Ah yes, that works too

Btw, in my solution should have done:

$\dfrac{\dfrac{1}{a}-\dfrac{1}{a+b}}{\dfrac{1}{b}-\dfrac{1}{a+b}} =$

$\dfrac{\dfrac{\dfrac{b}{a}}{a+b}}{\dfrac{\dfrac{a} {b}}{a+b}}} =$

$\dfrac{\dfrac{b}{a}}{a+b}*\dfrac{a+b}{\dfrac{a}{b} }=$

This part different

$\dfrac{\dfrac{b}{a}}{1}*\dfrac{1}{\dfrac{a}{b}}=$

$\dfrac{\dfrac{b}{a}}{\dfrac{a}{b}}=$

$\dfrac{b}{a}*{\dfrac{b}{a}=$

$\dfrac{{b}^{2}}{{a}^{2}}$