The equation $\frac{x-1}{x+1}+\frac{x+1}{x-1}=\frac{10}{3}$, where $x$ ≠ 1 and $x$ ≠ −1, reduces to which quadratic equation?

Put the left side over the common denominator $(x+1)(x-1)$: the numerator is $(x-1{)}^2+(x+1{)}^2=2x^2+2$, because the $-2x$ and +$2x$ cancel. So $\frac{2x^2+2}{x^2-1}=\frac{10}{3}$ → $3(2x^2+2)=10(x^2-1)$ → $6x^2+6=10x^2-10$ → $4x^2=16$ → $x^2-4=0$, giving $x$ = ±2. Check $x=2:\frac{1}{3}+3=\frac{10}{3}$. Option 1 mis-signs the 10 as it crosses over and would have no real roots at all; option 3 divides $4x^2=16$ by 16 instead of by 4; option 4 comes from missing the cancellation and expanding the numerator as $2x^2-2x+2$.