How do you solve #(x+1)/(x1)=2/(2x1)+2/(x1)#?
Now, I'm sure, figuring out the solution will be simple.
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To solve the equation (x+1)/(x1) = 2/(2x1) + 2/(x1), we can follow these steps:

Start by finding a common denominator for the fractions on the right side of the equation. The common denominator will be (x1)(2x1).

Multiply both sides of the equation by the common denominator to eliminate the fractions.

Simplify the equation by distributing and combining like terms.

Solve for x by isolating the variable on one side of the equation.

Check the solution by substituting the value of x back into the original equation to ensure it satisfies the equation.
Here are the steps in detail:

The common denominator is (x1)(2x1).

Multiply both sides of the equation by (x1)(2x1):
(x+1)(x1)(2x1) = 2(x1)(x1) + 2(2x1)(x1)
 Simplify the equation:
(x+1)(2x1) = 2(x1)^2 + 2(2x1)(x1)
Expand and simplify both sides:
2x^2  x + 2x  1 = 2(x^2  2x + 1) + 2(2x^2  3x + 1)
 Continue simplifying:
2x^2 + x  1 = 2x^2  4x + 2 + 4x^2  6x + 2
Combine like terms:
2x^2 + x  1 = 6x^2  6x + 4
 Rearrange the equation to have all terms on one side:
0 = 6x^2  2x^2 + 6x  x + 4 + 1
Combine like terms:
0 = 4x^2 + 5x + 5
 Set the equation equal to zero and attempt to factor or use the quadratic formula to solve for x. However, in this case, the equation cannot be factored easily, so we will use the quadratic formula:
x = (b ± √(b^2  4ac)) / (2a)
For our equation, a = 4, b = 5, and c = 5.
Substitute these values into the quadratic formula:
x = (5 ± √(5^2  4(4)(5))) / (2(4))
Simplify:
x = (5 ± √(25  80)) / 8
x = (5 ± √(55)) / 8
Since we have a negative value under the square root, the equation has no real solutions. Therefore, there are no values of x that satisfy the original equation.
Thus, the solution to the equation (x+1)/(x1) = 2/(2x1) + 2/(x1) is no solution.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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