can you show how to solve for x
x+1/x + 6/2x-3 =0
You need to use parentheses to clarify whether that means
(x+1)/x + 6/(2x-3) =0
or
(x+1)/x + (6/2x) + 3
or
x+ (1/x) + (6/2x)-3 =0
or
x+ (1/x) + 6/(2x-3) =0
The answers will be different in each case. Some will sing up as cubic equations.
In each case, get rid of the fractions by multiplying by common denomimators like x and (2x-3)
If the first equation is what you meant, you could write
(x+1)(2x+3) + 6(x+1) = 0
which becomes
2x^2 + 5x + 3 + 6x + 6 = 0
2x^2 +11x +9 = 0
(2x +9)(x+1)
x = -1 or -9/2