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1/a^2-1/a=6
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\[\huge \frac{1}{a^2}-\frac{1}{a}=6\]?
yes how do you solve that
you could multiply both sides by \(a^2\) as a good first step either that, or you have to do the subtraction
if you multiply both sides by \(a^2\) you get \[\huge 1-a=6a^2\] and now you have a quadratic equation to solve
if you do the subtraction first you get \[\frac{1-a}{a^2}=6\] then multiply both sides by \(a^2\) and you still get \[1-a=a^2\] so i guess it makes no difference solve \[6a^2+a-1=0\] for \(a\)
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