Quadratic Equation - Why is this wrong?

Question

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?

Answer 1

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?

Answer 2

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