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

View step-by-step help

Home

Math Problems

Calculus

Find higher derivatives of rational and radical functions

Full solution

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

Identify common denominator: First, let's identify the common denominator for the fractions on the left side of the equation. The common denominator will be the product of $(x+1)$ and $(x-1)$ , which is $(x^2 - 1)$ .
Rewrite fractions with common denominator: Now, let's rewrite each fraction on the left side of the equation with the common denominator $x^2 - 1$ . $\frac{2}{x+1}$ becomes $\frac{2(x-1)}{x^2 - 1}$ and $\frac{x}{x-1}$ becomes $\frac{x(x+1)}{x^2 - 1}$ .
Add fractions: Next, we add the two fractions on the left side of the equation. $\newline$ $(2)(x-1)/(x^2 - 1) + (x)(x+1)/(x^2 - 1) = (2x - 2 + x^2 + x)/(x^2 - 1)$
Simplify combined fraction: Simplify the numerator of the combined fraction. $\newline$ $(2x - 2 + x^2 + x)$ simplifies to $(x^2 + 3x - 2)$ . $\newline$ So, we have $\frac{x^2 + 3x - 2}{x^2 - 1}$ on the left side of the equation.
Equate left and right side: Now, we can equate the simplified left side of the equation to the right side of the equation, which is already given as $(2)/(x^2 - 1)$ . $\newline$ $(x^2 + 3x - 2)/(x^2 - 1) = (2)/(x^2 - 1)$
Equate numerators: Since the denominators are the same, we can equate the numerators. $x^2 + 3x - 2 = 2$
Subtract to set to zero: Subtract $2$ from both sides to set the equation to zero. $\newline$ $x^2 + 3x - 2 - 2 = 0$ $\newline$ $x^2 + 3x - 4 = 0$
Solve quadratic equation: Now, we need to solve the quadratic equation $x^2 + 3x - 4 = 0$ . We can do this by factoring, completing the square, or using the quadratic formula. Let's try factoring first. $\newline$ $(x + 4)(x - 1) = 0$