# Math Help - Algebraic Fractions Questions

1. ## Algebraic Fractions Questions

Hello everyone.

I have two urgent maths questions related to simplifying and solving algebraic fractions. I will be very gratified if you could pinpoint my mistakes and correct them with clear explanations.

1. $\dfrac{(a-b)^2 +a(a+b) -11b^2}{a^2 -4b^2}$

2. $(\dfrac{-2}{xy})^2$ X $\dfrac{y(y -x)^2}{4}$ X $\dfrac{x^2y^2}{(x^2 -y^2)}$

Here are my workings respectively-

1. $\dfrac{(a-b)^2 +a(a+b) -11b^2}{a^2 -4b^2}$

= $\dfrac{a^2-2ab+b^2+a^2+ab-11b^2}{(a-2b)(a+2b)}$

= $\dfrac{a^2+a^2-2ab+ab+b^2-11b^2}{(a-2b)(a+2b)}$

= $\dfrac{2a^2-ab-10b^2}{(a-2b)(a+2b)}$

= $\dfrac{(-2a+5b)(-a-2b)}{(a-2b)(a+2b)}$ ????

2. $(\dfrac{-2}{xy})^2$ X $\dfrac{y(y -x)^2}{4}$ X $\dfrac{x^2y^2}{(x^2 -y^2)}$

= $\dfrac{-4}{x^2y^2}$ X $\dfrac{y(y^2-2xy+x^2)}{4}$ X $\dfrac{x^2y^2}{(x-y)(x+y}$ ?????

2. 1) (-2a+5b)(-a-2b)=(2a-5b)(a+2b) now cancel (a+2b) from numerator and denominator then you will get the final answer
2) hints
(a/b)x(b/c)=a/c [b get cancel from numerator and denominator]
(-2/xy)^2=4/(xy)^2
(y-x)^2=(x-y)^2 use this to cancel (x-y) from denominator and numerator

3. Thank you nikhil, however I still could not figure out what you are trying to teach. Can you further elaborate on it?

4. For the first one:

$=\dfrac{(-2a+5b)(-a-2b)}{(a-2b)(a+2b)}$

This is the same as:

$=\dfrac{-1(-2a+5b)(a+2b)}{(a-2b)(a+2b)}$

The a + 2b cancel eachother out, leaving:

$=\dfrac{-1(-2a+5b)}{(a-2b)}$

$=\dfrac{2a-5b}{a-2b}$

For the second one:

$\dfrac{-4}{x^2y^2}\times \dfrac{y(y^2-2xy+x^2)}{4}\times \dfrac{x^2y^2}{(x-y)(x+y)}$

What nikhil was trying to say was that the x^2 * y^2 cancel eachother out because one is on the numerator and the other is on the denominator, leaving:

$\dfrac{y(y^2-2xy+x^2)}{4}\times \dfrac{4}{(x-y)(x+y)}~~~~~ = \dfrac{y(x-y)(x-y)}{4}\times \dfrac{4}{(x-y)(x+y)}$

Note: (-2)^2=4
Can you carry on from there?

5. Wow, that's genuinely quick!

So, for question 2:

$
\dfrac{y(y^2-2xy+x^2)}{4}\times \dfrac{4}{(x-y)(x+y)}~~~~~ = \dfrac{y(x-y)(x-y)}{4}\times \dfrac{4}{(x-y)(x+y)}

$

(cancel off the fours and then cancel off $(x-y)$ and $(x-y)$)

= $\dfrac{y(x-y)}{x+y}$

6. Seems right to me.

7. Hello, PythagorasNeophyte!

$1.\;\;\dfrac{(a-b)^2 +a(a+b) -11b^2}{a^2 -4b^2}$

$\text{Here are my workings:}$

$\dfrac{(a-b)^2 +a(a+b) -11b^2}{a^2 -4b^2}$

. . $=\; \dfrac{a^2-2ab+b^2+a^2+ab-11b^2}{(a-2b)(a+2b)}$

. . $=\;\dfrac{a^2+a^2-2ab+ab+b^2-11b^2}{(a-2b)(a+2b)}$

. . $=\;\dfrac{2a^2-ab-10b^2}{(a-2b)(a+2b)}$

. . $=\;\dfrac{(-2a+5b)(-a-2b)}{(a-2b)(a+2b)}$ .
?

Why did you factor like that?

If I asked you do reduce $\frac{6}{15}$, would you do this?

. . . . . . $\dfrac{6}{15} \;=\;\dfrac{(\text{-}2)(\text{-}3)}{(3)(5)}$

. . then say "Now what do I do?"