Another problem I need help with, also tell me name of the equation please

Solve for X:
((x+3)/(x-2)) - (2/x) = ((9x-8)/(x^2-2x))

All are fractions. Someone said this is a hyperbola, but hyperbola isn't taught in my class, I take algebra.

Factor the x^2 - 2x denominator on the right first:

[(x + 3) / (x - 2)] - (2 / x) = [(9x - 8) / x(x - 2)]

Clear the (x - 2) term in the denominators by multiplying through by (x - 2):

(x + 3) - [2(x - 2) / x] = [(9x - 8) / x]

Clear the x in the denominators by multiplying through by x:

x(x + 3) - 2(x - 2) = 9x - 8

Multiply out the parentheses:

x^2 + 3x - 2x + 4 = 9x - 8 (Note: With that x^2 term in it, it's called a quadratic equation.)

Combine like terms:

x^2 + x + 4 = 9x - 8

Subtract 9x from both sides and add 8 to both sides:

x^2 - 8x + 12 = 0

Factor to:

(x - 2)(x - 6) = 0

x = 6 (see below)


[added]
Good point in the previous post. x = 2 doesn't work in the original equation since it makes the (x - 2) denominators zero.

warning warning warning...

x = 2 is extraneous --
it does NOT work in the original equation because it makes denominators 0 (which is a big no-no)

always make sure to check solutions in rational (or radical) equations to make sure they work in the original equations

((x+3)/(x-2)) - (2/x) = ((9x-8)/(x^2-2x))
[x^2+3x-2x+4]/[x(x-2)] = [9x-8]/[x(x-2)]
x^2-8x+12=0
(x-6)(x-2)=0
x = 6 or 2
x cannot be 2 to avoid division by zero
x = 6
It is a quadratic equation.

Hyperbolas are curves in the Cartesian plane. You will learn it in Algebra eventually.