please help me solve this equation x/x-6 + 2/x = 1/x-6?

x/x-6+2/x=1/x-6
[x/x-6+2/x=1/x-6](x-6)x
x^2+2(x-6)=x
x^2+2x-12=x
x^2+2x-x-12=0
x^2+x-12=0
(x+4)(x-3)=0
x+4=0 x-3=0
x=-4 x=3 answer @(^_^)@

checking

-4/-4-6+2/-4=1/-4-6
-4/-10+2/-4=1/-10
-1/10=-1/10 my solution is correct!

2/(x-a million) - x/(x+3) = 6/(x²+2x-3) first of all, ingredient the quadratic. you're searching for 2 numbers that upload to +2, yet MULTIPLY to -3. What approximately 3 and -a million? 2/(x-a million) - x/(x+3) = 6/[(x+3)(x-a million)] which will supply you a splash which you extremely choose the denominator (backside) of all the fractions to be (x+3)(x-a million). the simplest way of doing that's to multiply one yet another fraction by potential of a few thing to grant you that. Multiply the 1st fraction by potential of (x+3)/(x+3) = a million, and the 2nd by potential of (x-a million)/(x-a million) = a million 2/(x-a million) * (x+3)/(x+3) - x/(x+3) * (x-a million)/(x-a million) = 6/[(x+3)(x-a million)] 2(x+3)/[(x+3)(x-a million)] - x(x-a million)/[(x+3)(x-a million)] = 6/[(x+3)(x-a million)] So you may now cancel the (x+3)(x-a million)! 2(x+3) - x(x-a million) = 6 2x + 6 - x² + x = 6 3x - x² = 0 x(3-x) = 0 2 numbers expanded jointly make 0. subsequently one or the different quantity could desire to be 0. x=0 or 3-x=0 x = 0 or x = 3 Do you remember the extensive difficulty we cancelled? this is a powerful concept to envision that that wasn't itself 0 (dividing by potential of 0 is a bad difficulty!) If x=0, (x+3)(x-a million) = (0+3)(0-a million) = -3 If x=3, (x+3)(x-a million) = (3+3)(3-a million) = 12 So, we are risk-free!

Remember that you and your sister should use brackets !

What you have given is in fact INCORRECT.

Should be shown as :-

x / (x - 6) + 2 / x = 1 / (x - 6)

x² + 2 (x - 6) = x

x² + x - 12 = 0

(x + 4) (x - 3) = 0

x = - 4 , x = 3

multiplying throughout by x(x-6).Then equation reduces to
x^2+2(x-6)=x
x^2+2x-12=x
x^2+x-12=0
(x+4)(x-3)=0
x= -4 or x=3

x/(x - 6) + 2/x = 1/(x - 6)
[x][x - 6][x/(x - 6) + 2/x] = [x][x - 6][1/(x - 6)]
x^2 + 2(x - 6) = x
x^2 + 2x - 12 = x
x^2 + 2x - x - 12 = 0
x^2 + x - 12 = 0
x^2 + 4x - 3x - 12 = 0
(x^2 + 4x) - (3x + 12) = 0
x(x + 4) - 3(x + 4) = 0
(x + 4)(x - 3) = 0

x + 4 = 0
x = -4

x - 3 = 0
x = 3

∴ x = -4 , 3

You have to get rid of the fractions by multiplying everything by x(x - 6) so you get x^2 + 2(x - 6) = x, then rearrange it a bit and solve the quadratic.

-6/x+x/2=x-6
-0.5x-6/x=-6
x+12/x=12
x^2+12=12x
x^2-12x+12=0
x=(-12+/-(48^(1/2)))/2

just tell her that you're a failure

x = -4 or +3