help!
2x/1+x-1/1-x=2
Get a common denominator. As I see your problem (you did not use parenthesis), the common denominator will be (1+x)(1-x)
Thank you!
To solve the equation 2x/(1+x) - 1/(1-x) = 2, we first need to find a common denominator for the two fractions. In this case, the common denominator will be (1+x)(1-x).
To do this, we multiply the first fraction by (1-x)/(1-x) and the second fraction by (1+x)/(1+x):
[2x/(1+x)] * [(1-x)/(1-x)] - [1/(1-x)] * [(1+x)/(1+x)] = 2
Simplifying, we have:
[2x(1-x)]/[(1+x)(1-x)] - (1+x)/(1+x)(1-x) = 2
Now we can combine the fractions over the common denominator:
[2x(1-x) - (1+x)]/[(1+x)(1-x)] = 2
Expanding and simplifying the numerator, we get:
[2x - 2x^2 - 1 - x] / [(1+x)(1-x)] = 2
Combining like terms in the numerator, we have:
[-2x^2 + x - 1] / [(1+x)(1-x)] = 2
Now, we can multiply both sides of the equation by (1+x)(1-x) to eliminate the denominator:
[-2x^2 + x - 1] = 2[(1+x)(1-x)]
Expanding the right side, we get:
-2x^2 + x - 1 = 2(1 - x^2)
Multiplying out the right side gives:
-2x^2 + x - 1 = 2 - 2x^2
Simplifying, we have:
x - 1 = 2
Adding 1 to both sides, we get:
x = 3
Therefore, the solution to the equation 2x/(1+x) - 1/(1-x) = 2 is x = 3.