Question

I need help understanding how to do this. I have a test tomorrow. I know I should know this but I am drawing blanks right now.

2/x+1-1/x=2x^2+x

Answer 1

\[\frac{2}{x+1}-\frac{1}{x}=2x^2+x\] like that ?

Answer 2

i hope not because i don't think this actually has any solutions
at least not real solutions

Answer 3

yes, ut the last one after the equal sign is 2/ x^2+x

Answer 4

oooh
\[\large \frac{2}{x+1}-\frac{1}{x}=\frac{2}{x^2+x}\]

Answer 5

yes

Answer 6

subtract on the left, the denominator will be \(x(x+1)=x^2+x\) same as on the right

Answer 7

so far that is the same answer i have so far aswell x(x+1)

Answer 8

\[\frac{2}{x+1}-\frac{1}{x}=\frac{2x-(x+1)}{x(x+1)}\]
\[=\frac{x-1}{x(x+1)}\]
now solve
\[\frac{x-1}{x(x+1)}=\frac{2}{x(x+1)}\] by
\[x-1=2\] etc

Answer 9

ok thanks