Paul solved the equation $$ \frac{2x^2-1}{x}=\frac{4x^2-1}{x}+6 $$ in the following steps:
(1) He simplified the fractions by dividing both the numerator and denominator by $x$: $$ 2x−1=4x−1+6 $$
(2) He moved all the variable terms to the left side of the equation and all the constants to the right side. Then, on each side of the equation, he combined like terms: $$\begin{aligned} 2x−4x&=−1+6+1 \cr −2x&=6 \end{aligned}$$
(3) He divided the equation by $-2$ to get the solution: $$\begin{aligned} \frac{−2x}{−2}&=\frac{6}{−2}\cr x&=−3 \end{aligned}$$
(4) Finally, he checked the solution to the equation: $$\begin{aligned} L &= \frac{2(−3)^2−1}{−3}=\frac{2 \cdot 9−1}{−3}=\frac{18−1}{−3}=−\frac{17}{3} \cr R &=\frac{4(−3)^2−1}{−3}+6=\frac{4 \cdot 9−1}{−3}+6=\frac{36−1}{−3} +6=\frac{−35}{3}+\frac{18}{3}=−\frac{17}{3} \end{aligned}$$
His classmates John, Erica, Peter, and Barbara commented on his solution. Which of them correctly commented on Paul's solution?
John says that Paul’s solution is incorrect and that Paul made an error in step (1).
Erica claims that Paul made an error in step (3). The negative sign on both sides of the equation leads to a positive solution, namely $x=3$.
Peter claims that Paul made an error in step (2). He should have got an equation $2x=6$.
Barbara thinks that Paul’s solution is correct. After all, the check came out.
From the assignment of the equation, it follows that $x$ must not be equal to zero. On this assumption, we can multiply both sides of the equation by the unknown $x$ (we do not forget to multiply the constant term $6$) and consecutively get: $$\begin{aligned} \frac{2x^2-1}{x}&=\frac{4x^2-1}{x}+6 \cr 2x^2-1&=4x^2-1+6x \cr 2x^2+6x&=0 \cr 2x(x+3)&=0 \end{aligned}$$ A product of factors is zero if and only if one or more factors are zero, i.e., $x=0$ or $x=−3$. As $x$ must not be equal to zero (otherwise, the fraction would not be defined), there is only one solution left, namely $x=−3$.