I don't understand how to simplify this fraction: $$\frac{12ab-12b}{8ac-24c}$$ This is my idea to what should be done, but I think it is totally wrong: $$\frac{12ab-12b}{8ac-24c}=\frac{12\cdot a\cdot b-12\cdot b}{8\cdot a\cdot c-24\cdot c}=\frac{12b^2-12b}{8c-24c}=\frac{b}{8c-24c}=?$$

Edit: I want to simplify the fraction.

  • 1
    $\begingroup$ Solve? What's to solve? Simplify, maybe. Note that you can cancel common factors of $4$ from the numerator and denominator. $\endgroup$ – J. M. isn't a mathematician Sep 10 '11 at 13:34
  • $\begingroup$ You want to solve this fraction means you want to simplify it? If you want to simply this fraction further the you can only divide the numerator and denominator by $4$. That's al. $\endgroup$ – user9413 Sep 10 '11 at 13:35
  • $\begingroup$ You don't solve an algebraic fraction, you simplify it. $\endgroup$ – Américo Tavares Sep 10 '11 at 13:36
  • $\begingroup$ Two important things: (1) You need to FACTOR the numerator and denominator in order to get things that will cancel. (2) You cannot cancel something that appears in just one term in the numerator or denominator, and is therefore not a factor of the numerator or the denominator, and that means you can't cancel the $a$ from $12ab$ and from $8ac$. (How you somehow got $12b^2-12b$ to be just $b$ I won't try to guess.) $\endgroup$ – Michael Hardy Sep 10 '11 at 15:23

In general

$$\frac{12ab-12b}{8ac-24c}\ne \frac{12b^2-12b}{8c-24c}.$$

You can divide $^1$ both numerator and denominator by $4=\gcd(12,8,24)$:


If you want you can factor the numerator and denominator

$$\frac{3ab-3b}{2ac-6c}=\frac{3b\left( a-1\right) }{2c\left( a-3\right) }.$$

Since there are no common factors you cannot simplify further.


$^1$ In greater detail:

$$\frac{12ab-12b}{8ac-24c}=\dfrac{\dfrac{12ab-12b}{4}}{\dfrac{8ac-24c}{4}}=\dfrac{ \dfrac{12ab}{4}-\dfrac{12b}{4}}{\dfrac{8ac}{4}-\dfrac{24c}{4}}=\dfrac{\dfrac{12}{4} ab-\dfrac{12}{4}b}{\dfrac{8}{4}ac-\dfrac{24}{4}c}=\dfrac{3ab-3b}{2ac-6c}.$$

  • $\begingroup$ Thanks, now it is much more clear to me. I simply tried to mix things that do not go together i.e. 12ab is different from 12b $\endgroup$ – esp78 Sep 10 '11 at 14:50
  • $\begingroup$ @esp78: Glad helping you. $\endgroup$ – Américo Tavares Sep 10 '11 at 15:03

I have several bits to say here.

Firstly, the phrase 'solve' implies an equation, something like $\text{expression} = \text{expression}$. Here, we just have an expression, so there is no equation to solve. I suspect you mean to 'simplify' the fraction.

Secondly, let's talk about what things are you allowed to do to simplify a fraction? You tried $$\frac{12ab-12b}{8ac-24c}=\frac{12*a*b-12*b}{8*a*c-24*c}=\frac{12b^2-12b}{8c-24c}=\frac{b}{8c-24c}=?$$ However, this is not correct. In particular, in the middle two expressions, I noticed that you cancelled the a's from the left numerator and left denominator. This would be like saying that $3 = \dfrac{8 + 7}{4 + 1} = \dfrac{2 + 7}{1 + 1} = \dfrac{9}{2}$. That's just not true. I'll return to this in my third point. Before that, I wanted to talk about one more problem: in the last two expressions, you wrote $12 b^2 - 12b = b$. This is also not correct. For a concrete example, we look at $12 \cdot 3^2 - 12 \cdot 3 = 72 \not = 3$. I'm a bit uncertain as to what to say to that, except that it's very not allowed. Though it's not useful here, I do mention that you could factor that expression: $12b^2 - 12b = 12b(b - 1)$. But that's really all there is to do there.

Thirdly, how should you actually proceed? One way is to factor the top and bottom as much as you can. For example, $12ab - 12b = 12b(a - 1)$ and $8ac - 24c = 8c(a - 3)$. Rewriting my main fraction, we get $\displaystyle \frac{12ab - 12b}{8ac - 24c} = \frac{12b(a-1)}{8c(a-3)}$. Then I note that we could rewrite $\dfrac{12}{8}$ as $\dfrac{3}{2}$. But there is no more simplification to be done.

Finally, I note that some people consider factoring to be a simplification. I do not, and I get tired when grading homework or tests when students arbitrarily start factoring expressions into things that I consider far less simple - but the main point is that different teachers have different expectations. Make sure you are aware of those.

  • $\begingroup$ @mixedmath: You're right that factoring, by itself, is not necessarily a simplification. But it can be used further in a process of simplification, when there are some common factors which can be canceled. And how could we know that there are some until we factor? ;) Also, factoring polynomial expressions allows us to see their roots (solutions), while the non-factored form shows us nothing much. $\endgroup$ – SasQ Nov 8 '13 at 18:23

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