Solving a rational equation?

Question

how do I solve


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

Answer 1

First get a common denominator on the right:
2 / (x-2) = x / (x-1) - 2(x - 1)/(x-1)

Now combine the fractions on the right:
2 / (x-2) = (x - 2(x - 1)) / (x - 1)

Simplify the numerator:
2 / (x-2) = (x - 2x + 2) / (x - 1)
2 / (x-2) = (-x + 2) / (x - 1)

Pull out a -1 on the right:
2 / (x-2) = -(x - 2) / (x - 1)

Multiply by (x-2) on both sides:
2 = -(x - 2)(x - 2) / (x - 1)

Multiply by (x-1) on both sides:
2(x - 1) = -(x - 2)(x - 2)

Now multiply out:
2x - 2 = -x^2 +4x - 4

Now put everything on one side:
x^2 -4x + 2x -2 + 4 = 0
x^2 - 2x + 2 = 0

Now use the quadratic formula to solve for x.
x = [2 +/- sqrt(4 - 4(2)) ] / 2
x = 1 +/- sqrt(-4)/2
x = 1 +/- i

Answer 2

I assume you mean

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

Multiply both sides by (x-1)(x-2), to get rid of the fractions:

2(x-1) = x(x-2) - 2(x-1)(x-2)

Multiply everything out:

2x-2 = x^2 - 2x - 2x^2 + 6x - 4

Combine like terms:

x^2 - 2x + 2 = 0

Quadratic formula:

x = 1 +/- i, where i = sqrt(-1), the imaginary unit.

Answer 3

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

2/(x-2) = -1 + 1/(x-1)
2 = (2-x) + (x-2)/(x-1)
2 = 3 -x - 1/(x-1)
x -1 = -1/(x-1)
we could say then if a = -1/a, then a = i or -i (obviously)
thus if x - 1 = i, x = i + 1
if x - 1 = -i, x = 1 - i

Answer 4

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

Multiply everything by (x - 1)(x - 2)

2(x - 1) = x(x - 2) - 2(x - 1)(x - 2)
2x - 2 = x^2 - 2x - 2(x^2 - 2x - x + 2)
2x - 2 = x^2 - 2x - 2(x^2 - 3x + 2)
2x - 2 = x^2 - 2x - 2x^2 + 6x - 4
2x - 2 = -x^2 + 4x - 4
x^2 - 2x + 2 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-2) ± sqrt((-2)^2 - 4(2)(1)))/(2(1))
x = (2 ± sqrt(4 - 8))/2
x = (2 ± sqrt(-4))/2
x = (2 ± 2i)/2)
x = 1 ± i

ANS :

x = 1 - i or 1 + i

Answer 5

Add 2 to both sides

2/X = X/X - 1

Times everything by X to get it out of the fractions

2 = X - X.....
2=0..... obviously not true.
Your equation is unsolvable.

Answer 6

... 10/(x-a million) + a million = 20/(x²-a million) ... x ? ±a million ? Else equation will become undefined or 10/(x-a million) + a million = 20/[ (x-a million)(x+a million) ] or 10(x+a million) + (x-a million)(x+a million) = 20 or 10x + 10 + x² - a million = 20 or x² + 10x - eleven = 0 or (x - a million) (x + eleven) = 0 or x = a million ....... ? Reject through preliminary constraint or x = -eleven .... ? very final answer

Answer 7

2/(x-2)=[x-2(x-1)]/(x-1)
2/(x-2)=[x-2x+2]/(x-1)
2/(x-2)=(-x+2)/(x-1)
2(x-1)=(x-2)((-x+2)
2x-2=-(x-2)^2
2x-2=-(x^2-4x+4)
2x-2=-x^2+4x-4
x^2-2x+2=0
x=[2+/-(4-8)]/2
=1+/-i