To solve the equation below, we factor the numerator and cancel:

$$\frac{c^2 + 6c -27}{c-3} +2c = 23 \color{DarkGreen} \implies\frac{(c-3)(c+9)}{c-3} +2c= 23 \color{DarkGreen}\implies c = \frac{14}{3}\neq 3$$ However, if I try to solve this by another method:
\begin{align}\frac{c^2 + 6c -27}{c-3} &= 23 - 2c \\[0.2cm] c^2 + 6c -27 &= (23 - 2c)(c-3)\\[0.2cm] 3c^2 -23c - 42 &= 0 \\[0.2cm] c &\neq \frac{14}{3}\end{align}

Then I get the wrong answer. Obviously, I am doing something wrong. But what?

  • $\begingroup$ That $-42$ is incorrect $\endgroup$ – imranfat Oct 19 '16 at 12:41
  • $\begingroup$ Can you elaborate. I have gone over the calculation a few times, but seem to be missing it. I am sure it is something obvious. $\endgroup$ – Starlight Oct 19 '16 at 12:43
  • $\begingroup$ @imranfat $42$ is never incorrect, it is the answer to life, the universe and everything... $\endgroup$ – Kevin Oct 19 '16 at 12:47
  • $\begingroup$ @Bacon Well, yes, but OP got $-42$ instead of $42$... $\endgroup$ – 5xum Oct 19 '16 at 12:48
  • $\begingroup$ @5xum Indeed, however to counter (and remain off topic!) once you know what the question actually is, you'll know what the answer means... $\endgroup$ – Kevin Oct 19 '16 at 12:55

The mistake happens between the line

$$\displaystyle c^2 + 6c -27 = (23 - 2c)(c-3)$$ and

$$\displaystyle 3c^2 -23c - 42 = 0$$

You made two mistakes. One, the obvious, is that the top and bottom equation are not equivalent.

However, the more important mistake you did was trying to rush this step. You did a lot of things at once, and that is a huge risk. So, slow down and do it step by step:

$$\begin{align}c^2+6c-27 &= 23c - 3\cdot 23 - 2c^2+6c\\ c^2+6c-27-23c+69+2c^2-6c&=0\\ 3c^2-23c+42 &= 0\end{align}$$

Not the same thing you got, right?


It should be $$3 c^2 - 23 c + 42 = 0$$ and everything is as expected.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.