# How to solve this algebraic fraction?

Could someone please help me with solving this algebraic fraction. I tried it a few times and I got the wrong answer all of the times. My brother also tried, who had recently finished Matric and he is very good at Maths. He also got the wrong answer. I'm completely lost with this one!

algebraic fraction

If you can't read the image (I apologize, I can't write neatly on paint on a computer!), here is a "worded fraction":

$$\frac{a^2 + ab}{2b - a}: \frac{a}{2a - 4b}$$

The answer we got from our teacher was $2(a - 2b)$

I'm sorry if this isn't clear enough, but any help would be very much appreciated!

PS I know how to factorise everything, simplify and such, I just can't figure this one out. I am a grade 9 student.

Seems that Wolframalpha and me don't agree with the answer $2(a-2b)$ of your teacher. We can observe that for $a \neq 0$ and $2b-a \neq 0$ holds:

$$\begin{array}{rcl}\frac{a^2+ab}{2b-a}:\frac{a}{2a-4b} &=& \frac{a^2+ab}{2b-a}\cdot \frac{2a-4b}{a}\\ &=& \frac{a^2+ab}{a}\cdot \frac{2a-4b}{2b-a} \\ &=& \frac{(a+b)a}{a}\cdot\frac{-2(2b-a)}{2b-a}\\ &=& -2(a+b)\end{array}$$

• Nice and simple answer - nothing more to say... Good job!!! – johannesvalks Jun 22 '14 at 18:34

You want to simplify $$\frac{\frac{a^2+ab}{2b-a}}{\frac{a}{2a-4b}}$$ Using $\frac{a}{b}=a(b)^{-1}$ for simplicity, this is $$((a^2+ab)(2b-a)^{-1})((a)(2a-4b)^{-1})^{-1}$$ or $$(a^2+ab)(2b-a)^{-1}(a)^{-1}(2a-4b)$$ Rewriting, $$((a^2+ab)(a)^{-1})((2b-a)^{-1}(2a-4b))$$ Now, $a^2+ab=a(a+b)$ and $(2a-4b)=-2(2b-a)$, so this is $$(a)(a^{-1})(a+b)(2b-a)^{-1}(2b-1)(-2)$$ Canceling, this is $$\boxed{(-2)(a+b)}$$

Note: I am working off the word version. This doesn't match what you have from you teacher. Did he mean $-2a-2b$?

Also, PS: this doesn't look much like linear algebra, especially if you're only in 9th grade. And if it is, just know that Vector Spaces let you do most algebraic manipulations as normal.

Highlighting some key passages with different colours, we have: \begin{align} \frac{\color{blue}{a^2+ab}}{2b-a}:\frac{a}{\color{green}{2a-4b}}&=\frac{\color{blue}{a(a+b)}}{2b-a}:\frac{a}{\color{green}{2(a-2b)}}=\\[2mm] &=\frac{a(a+b)}{\color{red}{2b-a}}\cdot\frac{2(a-2b)}{a}=\\[2mm] &=\color{red}{-}\frac{a(a+b)}{\color{red}{a-2b}}\cdot\frac{2(a-2b)}{a}=\\[2mm] &=-2\cdot\frac{a}{a}\cdot\frac{a-2b}{a-2b}\cdot(a+b)=\\[2mm] &=-2(a+b). \end{align} This holds for all $$a,b\in\mathbb{R}$$ such that $$0\ne a\ne 2b$$.