x/x-2 + x-1/x+1 = -1
I am almost certain you mean
x/(x-2) + (x-1)/(x+1) = -1
the LCD is (x-2)(x+1) , so multiply each term by that to get
x(x+1) + (x-1)(x-2) = -1(x-2)(x+1)
expand, re-arrange to get a quadratic.
Then solve using your favourite method.
wait what do you mean your favorite method?
Methods to solve quadratics:
1. by factoring, will not always work
2. by completing the square, best way if coefficient of the square term is 1 and the middle term is even
3. using the quadratic formula, sure-proof way
To solve the equation, we need to find the value of x that satisfies the equation. Let's go step by step:
1. Start by finding a common denominator for the fractions on the left side of the equation. In this case, the common denominator is (x - 2)(x + 1):
x/(x - 2) + (x - 1)/(x + 1) = -1
(x(x + 1) + (x - 1)(x - 2))/(x - 2)(x + 1) = -1
2. Now, simplify the numerator:
(x^2 + x + x^2 - 3x + 2)/(x - 2)(x + 1) = -1
(2x^2 - 2x + 2)/(x - 2)(x + 1) = -1
3. Multiply both sides of the equation by the common denominator to eliminate the fractions:
(2x^2 - 2x + 2)/(x - 2)(x + 1) * (x - 2)(x + 1) = -1 * (x - 2)(x + 1)
2x^2 - 2x + 2 = -1 * (x - 2)(x + 1)
2x^2 - 2x + 2 = -(x^2 - x - 2)
4. Distribute the negative sign on the right side:
2x^2 - 2x + 2 = -x^2 + x + 2
5. Rearrange the equation by combining like terms:
2x^2 - 2x + 2 = -x^2 + x + 2
2x^2 + x^2 - 2x - x + 2 - 2 = 0
3x^2 - 3x = 0
3x(x - 1) = 0
6. Set each factor equal to zero and solve for x:
3x = 0 or x - 1 = 0
x = 0 or x = 1
7. Therefore, the values of x that satisfy the equation are x = 0 and x = 1.
To verify the solutions, substitute them back into the original equation and see if it holds true.