1

Currently doing high school math and can't get this one right. I think I'm using entirely incorrect practices and am trying to pinpoint what it is. Could someone tell me where exactly I went wrong?

Original expression: $$\left(\frac{b}{a^{2}+ab}-\frac{b-a}{b^2+ab}\right)\times \left(\frac{a^2}{b^3-a^2b}+\frac{1}{a+b}\right)^{-1}$$

When I simplify the equations in parentheses I get: $$\frac{b^2-ab+a^2}{ab(a+b)}\times\frac{1}{a^{2}}$$

Which should result in: $$\frac{b^2+a(a-b)}{a^3b(a+b)}=\frac{b^2-a}{a^3b}$$

That is of course not right. The correct answer according to the book is $\frac{b-a}{a}$

Zae
  • 17
  • Your last equality is false. it is the low school maths. – hamam_Abdallah Dec 15 '18 at 20:59
  • Your derivation of the second factor is incorrect. You should find $\frac{a^3+b^3}{(a+b)(b^3-a^2b)}=\frac{a^2-ab+b^2}{b(b+a)(b-a)}$. So the $a^2-ab+b^2$ and $b(b+a)$ will cancel out when you make the division and you will get the final result. – Aphelli Dec 15 '18 at 21:04
  • Thank you so so much @Mindlack this solved it easily! – Zae Dec 15 '18 at 21:34

1 Answers1

0

Go slowly. Do it in blocks.

First block: \begin{align} \frac{b}{a^{2}+ab}-\frac{b-a}{b^2+ab} &=\frac{b}{a(a+b)}-\frac{b-a}{b(a+b)} \\[4px] &=\frac{b^2-a(b-a)}{ab(a+b)} \\[4px] &=\frac{a^2-ab+b^2}{ab(a+b)} \end{align}

Second block: \begin{align} \frac{a^2}{b^3-a^2b}+\frac{1}{a+b} &=\frac{a^2}{b(b^2-a^2)}+\frac{1}{a+b} \\[4px] &=\frac{a^2}{b(b-a)(b+a)}+\frac{1}{a+b} \\[4px] &=\frac{a^2+b(b-a)}{b(b-a)(b+a)} \\[4px] &=\frac{a^2-ab+b^2}{b(b-a)(b+a)} \end{align}

Putting the pieces together: $$ \left(\frac{b}{a^{2}+ab}-\frac{b-a}{b^2+ab}\right)\times \left(\frac{a^2}{b^3-a^2b}+\frac{1}{a+b}\right)^{-1}= \frac{a^2-ab+b^2}{ab(a+b)}\frac{b(b-a)(b+a)}{a^2-ab+b^2}= \frac{b-a}{a} $$ Now you see that you managed the first block correctly, but failed in the second one. Check carefully your steps and you'll see where you went astray.

egreg
  • 238,574
  • Thank you so much for taking the time to write it all out! A comment already pointed out my mistake, and this illustrates everything perfectly. – Zae Dec 16 '18 at 09:58