2x/(x^2-2x-3)+ 3x/(x^2+2x-3)=7/8

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To solve this rational equation, we first need to find a common denominator for the two fractions on the left side of the equation.

Given that the denominators are quadratic expressions, let's factor them first:

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

Now, we can rewrite the equation with a common denominator:

2x/(x - 3)(x + 1) + 3x/(x - 1)(x + 3) = 7/8

Next, we can multiply each fraction by the missing factor in the denominator:

2x(x - 1)/(x - 3)(x + 1)(x - 1) + 3x(x - 3)/(x - 1)(x + 3)(x - 3) = 7/8

Now, the denominators are the same, and we can combine the fractions:

[2x^2 - 2x + 3x^2 - 9x]/[(x - 3)(x + 1)(x - 1)] = 7/8

Combine like terms in the numerator:

[5x^2 - 11x]/[(x - 3)(x + 1)(x - 1)] = 7/8

Now, we have a single fraction on both sides of the equation. To solve for x, we can cross multiply:

8(5x^2 - 11x) = 7[(x - 3)(x + 1)(x - 1)]

Expand and simplify both sides of the equation:

40x^2 - 88x = 7(x^3 - x^2 - 3x + 3)

40x^2 - 88x = 7x^3 - 7x^2 - 21x + 21

Rearrange the terms to set the equation equal to zero:

7x^3 - 47x^2 + 67x - 21 = 0

Now, the equation is in standard form, and we can use methods such as factoring, synthetic division, or a numerical method to solve for x.