Question

6x^-2 -7x^-1 +1=0 Solve by making appropriate substitutions

Answer 1

Let $y=x^{-1}$, so that $y^2=x^{-2}$. Then

$6x^{-2}-7x^{-1}+1=0\iff6y^2-7y+1=0\implies (6y-1)(y-1)=0$

which has two solutions, $y=\dfrac16$ and $y=1$. So the solutions in terms of $x$ would be

$x^{-1}=\dfrac16\implies x=6$
$x^{-1}=1\implies x=1$