{:[(2)/(x+1)+(x)/(x-1)=(2)/(x^(2)-1)]:}

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Write Equations: First, let's write down the system of equations clearly: (1) $\frac{2}{x+1} + \frac{x}{x-1} = \frac{2}{x^2-1}$ (2) $\frac{x}{x-1} = \frac{2}{(x-1)(x+1)} - \frac{2}{x+1}$ We need to solve this system to find the value of x.Factorize and Simplify: Notice that $x^2 - 1$ is the difference of squares and can be factored into $(x+1)(x-1)$. This means that the right side of equation (1) is already simplified. We can use this information to simplify the left side of equation (1) by finding a common denominator.Combine Terms: To find a common denominator for the left side of equation (1), we multiply the first term by $\frac{x-1}{x-1}$ and the second term by $\frac{x+1}{x+1}$ to get: $\frac{2(x-1)}{(x+1)(x-1)} + \frac{x(x+1)}{(x-1)(x+1)}$Expand Numerator: Now, we combine the terms over the common denominator: $\frac{2(x-1) + x(x+1)}{(x+1)(x-1)}$Combine Like Terms: Next, we expand the numerator: $2x - 2 + x^2 + x$Set Numerators Equal: Combine like terms in the numerator: $x^2 + 3x - 2$Solve Quadratic Equation: Now we have a single fraction on the left side of equation (1): $\frac{x^2 + 3x - 2}{(x+1)(x-1)} = \frac{2}{(x+1)(x-1)}$Check Solutions: Since the denominators are the same, we can set the numerators equal to each other: $x^2 + 3x - 2 = 2$Valid Solution: Subtract 2 from both sides to set the equation to zero: $x^2 + 3x - 4 = 0$Invalid Solution: Now we need to solve the quadratic equation $x^2 + 3x - 4 = 0$. This can be factored into: $(x + 4)(x - 1) = 0$Setting each factor equal to zero gives us the possible solutions for x: $x + 4 = 0$ or $x - 1 = 0$Solving each equation for x gives us: $x = -4$ or $x = 1$However, we must check these solutions against the original equations to ensure they do not make any denominators zero. Plugging $x = -4$ into the denominators of the original equations, we see that none of the denominators become zero, so $x = -4$ is a valid solution.Plugging $x = 1$ into the denominators of the original equations, we see that $x = 1$ would make the denominators of both equations zero, which is not allowed. Therefore, $x = 1$ is not a valid solution.

$\frac{2}{x+1}+\frac{x}{x-1}=\frac{2}{x^{2}-1}$

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2x+1+xx−1=2x2−1 \frac{2}{x+1}+\frac{x}{x-1}=\frac{2}{x^{2}-1}x+12​+x−1x​=x2−12​

$\frac{2}{x+1}+\frac{x}{x-1}=\frac{2}{x^{2}-1}$