Two days ago "Steve" helped me with this problem. I am grateful to him. I didn't get it and asked for him to help me again. He showed it to me again, but I needed more steps shown. Then I think he was finished for tutoring for the day before I could see it again with more steps. I understand factoring out the 1/3 on the left hand side and I understand factoring the denominator on the right hand side. After that I have problems. Please help me. The problem is: (x/(3x+6))- ((x+1)/(3x-6))=1/((x^2)-4)

Thank you.

To solve the equation:

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

Let's first simplify the equation by factoring out the common factors on each side.

On the left side of the equation, we can factor out 1/3 from both terms:

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

Now the equation becomes:

(x/3)(1/(x+2)) - (x+1)/(3(x-2)) = 1/(3(x+2)(x-2))

Next, let's find a common denominator for the two fractions on the left side.

The common denominator between (x+2) and (x-2) is (x+2)(x-2), so we need to multiply and divide the first fraction by (x-2), and the second fraction by (x+2) to get the same denominator:

[(x/3)(1/(x+2))(x-2)]/[(x+2)(x-2)] - [(x+2)(x+1)]/[(3(x-2))(x+2)(x-2)] = 1/(3(x+2)(x-2))

Simplifying further:

(x(x-2))/(3(x+2)(x-2)) - [(x+2)(x+1)]/[(3(x-2))(x+2)(x-2)] = 1/(3(x+2)(x-2))

Canceling out the common factors:

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

Expanding the numerator of the second fraction:

x - (x^2 + 3x + 2) = 1

Simplifying:

x - x^2 - 3x - 2 = 1

Combining like terms:

- x^2 - 2x - 3 = 1

Rearranging the equation:

x^2 + 2x + 4 = 0

Now, we have a quadratic equation that we can solve. We can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For this equation, a = 1, b = 2, and c = 4.

x = (-2 ± √(2^2 - 4(1)(4))) / (2(1))

Simplifying:

x = (-2 ± √(4 - 16)) / 2

x = (-2 ± √(-12)) / 2

Since the square root of a negative number results in an imaginary number, there are no real solutions for x in this equation.

Hence, the given equation has no solutions in the real number system.