On the weekend I help out maths at a club for students of 15-16 years of age. I did a survey on a particular algebra question. I asked all of them roughly 200 students(not in a sitting but spread out the entire days) to simplify $\color{red}{4-{{{{x-2\over x+2}}}}}$ into a single fraction. I collected all the answers and counted them.

I found out that 70% of the students got it wrong!

They gave $\color{blue}{3x+6\over x+2}$ instead of $\color{green}{3x+10\over x+2}$

Where do you think they went wrong? What advice of teaching techniques do you advice tutors to do, to help teach them better, so in the future they won't get it wrong again in these kind of basic algebraic fractions questions?

  • 3
    $\begingroup$ Well, I think all had the same idea: write $4 = \frac{4(x+2)}{x+2}$. The main problem may lie in calculating the enumerator correctly: $4(x+2)- \color{red}{(}x-2\color{red}{)}$. They simply forgot to put the brackets. $\endgroup$ – Niklas Feb 5 '17 at 17:27
  • $\begingroup$ Agree. They probably miss the double negative of $-(x-2)$. If they write the whole expression with brackets, it will be much clearer. $\endgroup$ – Mythomorphic Feb 5 '17 at 17:29
  • $\begingroup$ Can you ask them again (different question probably) and ask them to write the answer on one sheet of paper and then look at what exactly they got wrong. Although I agree with what @user8795 wrote, I feel it is (and any other answer will be) just speculation. $\endgroup$ – Jan Feb 5 '17 at 17:31
  • $\begingroup$ We can all learn from our mistakes. Sign mistakes are easy to make, so students should learn from them early and often. Checking answers (as @Yves suggests, by evaluation) is worth teaching. $\endgroup$ – hardmath Feb 5 '17 at 18:15
  1. Tell them to decompose the computation

$$4-\frac{x-2}{x+2}=\frac{4(x+2)-(x-2)}{x+2}$$ instead of rushing to


  1. Make them check with numerical instances.

With $x=2$, $4$ vs. $\dfrac{12}4$.


Possibly done mistake like: $${4-{x-2\over x+2}} = \frac{4x+8-x-2}{x+2} = \frac{3x+6}{x+2}$$

Instead of $${4-{x-2\over x+2}} = \frac{4x+8-x+2}{x+2} = \frac{3x+10}{x+2}$$


Distributing the negative sign to the numerator before doing addition would help


  • $\begingroup$ I like your idea, removing the negative sign first $\endgroup$ – gymbvghjkgkjkhgfkl Feb 5 '17 at 18:01

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