We can start by finding a common denominator for the two fractions on the left side of the equation:
x/(x-2) + (x-1)/(x+1) = 1
LCD = (x-2)(x+1)
We can then rewrite each fraction with the LCD as the denominator:
x(x+1)/[(x-2)(x+1)] + (x-1)(x-2)/[(x-2)(x+1)] = 1
Simplifying the fractions:
[x(x+1) + (x-1)(x-2)]/[(x-2)(x+1)] = 1
Expanding the brackets:
[x^2 + x + x^2 - 3x + 2]/[(x-2)(x+1)] = 1
Simplifying the numerator:
[2x^2 - 2x + 2]/[(x-2)(x+1)] = 1
Multiplying both sides by the denominator:
2x^2 - 2x + 2 = (x-2)(x+1)
Expanding the right side:
2x^2 - 2x + 2 = x^2 - x - 2
Bringing all the terms to one side:
x^2 - x - 4 = 0
Using the quadratic formula to solve for x:
x = (1 ± √17)/2
Checking our solutions:
x = (1 + √17)/2:
x/(x-2) + (x-1)/(x+1) = (1 + √17)/2 / ([1 + √17]/2 - 2) + [(1 + √17)/2 - 1]/([1 + √17]/2 + 1) = 1
x = (1 - √17)/2:
x/(x-2) + (x-1)/(x+1) = (1 - √17)/2 / ([1 - √17]/2 - 2) + [(1 - √17)/2 - 1]/([1 - √17]/2 + 1) = 1
Therefore, our solutions are x = (1 ± √17)/2.
x/x-2 + x-1/x+1=1
5 answers
sorry - you messed up here
2x^2 - 2x + 2 = x^2 - x - 2
it should be
x^2 - x + 4 = 0
now the discriminant is negative, so there are no real solutions.
x = (1 + √15 i)/2
2x^2 - 2x + 2 = x^2 - x - 2
it should be
x^2 - x + 4 = 0
now the discriminant is negative, so there are no real solutions.
x = (1 + √15 i)/2
You are correct! Thank you for catching the mistake. The correct solution is x = (1 + √15 i)/2 and x = (1 - √15 i)/2.
I think this problem was originally posted as
x/(x-2) + (x-1)/(x+1) = -1
so the solutions are x=0 or x=1
x/(x-2) + (x-1)/(x+1) = -1
so the solutions are x=0 or x=1
Thank you for pointing that out. You are correct that if the original problem was x/(x-2) + (x-1)/(x+1) = -1, then the solutions would be x=0 or x=1.
We can start by finding a common denominator for the two fractions on the left side of the equation:
x/(x-2) + (x-1)/(x+1) = -1
LCD = (x-2)(x+1)
We can then rewrite each fraction with the LCD as the denominator:
x(x+1)/[(x-2)(x+1)] + (x-1)(x-2)/[(x-2)(x+1)] = -1
Simplifying the fractions:
[x(x+1) + (x-1)(x-2)]/[(x-2)(x+1)] = -1
Expanding the brackets:
[x^2 + x + x^2 - 3x + 2]/[(x-2)(x+1)] = -1
Simplifying the numerator:
[2x^2 - 2x + 2]/[(x-2)(x+1)] = -1
Multiplying both sides by the denominator:
2x^2 - 2x + 2 = -(x-2)(x+1)
Expanding the right side:
2x^2 - 2x + 2 = -x^2 + x + 2
Bringing all the terms to one side:
3x^2 - 3x = 0
Factoring out x:
3x(x-1) = 0
Therefore, our solutions are x=0 or x=1.
We can start by finding a common denominator for the two fractions on the left side of the equation:
x/(x-2) + (x-1)/(x+1) = -1
LCD = (x-2)(x+1)
We can then rewrite each fraction with the LCD as the denominator:
x(x+1)/[(x-2)(x+1)] + (x-1)(x-2)/[(x-2)(x+1)] = -1
Simplifying the fractions:
[x(x+1) + (x-1)(x-2)]/[(x-2)(x+1)] = -1
Expanding the brackets:
[x^2 + x + x^2 - 3x + 2]/[(x-2)(x+1)] = -1
Simplifying the numerator:
[2x^2 - 2x + 2]/[(x-2)(x+1)] = -1
Multiplying both sides by the denominator:
2x^2 - 2x + 2 = -(x-2)(x+1)
Expanding the right side:
2x^2 - 2x + 2 = -x^2 + x + 2
Bringing all the terms to one side:
3x^2 - 3x = 0
Factoring out x:
3x(x-1) = 0
Therefore, our solutions are x=0 or x=1.