Solve for $x$.
$\frac{3}{x-1}-\frac{1}{x^2-1}=\frac{5}{x-1}$
To solve the equation \(\frac{3}{x-1} - \frac{1}{x^2-1} = \frac{5}{x-1}\), we first analyze the expression to determine the valid domain for \(x\). The denominators of the fractions must not be zero, so we require \(x-1 eq 0\) and \(x^2-1 eq 0\). This implies that \(x eq 1\) and \(x eq -1\). Therefore, the domain of the equation is all real numbers except for \(1\) and \(-1\).
We can simplify the equation by grouping terms with the same denominator. Subtract \(\frac{3}{x-1}\) from both sides to isolate the terms:
\[-\frac{1}{x^2-1} = \frac{5}{x-1} - \frac{3}{x-1}\]
This simplifies to:
\[-\frac{1}{x^2-1} = \frac{2}{x-1}\]
Recall that \(x^2-1 = (x-1)(x+1)\). Substituting this into our equation, we get:
\[-\frac{1}{(x-1)(x+1)} = \frac{2}{x-1}\]
Since \(x
eq 1\), we can multiply both sides of the equation by \((x-1)\) to eliminate it from the denominators:
\[-\frac{1}{x+1} = 2\]
Multiplying both sides by \((x+1)\) gives:
\[-1 = 2(x+1)\]
\[-1 = 2x + 2\]
Subtracting 2 from both sides results in:
\[-3 = 2x\]
Dividing by 2, we find:
\[x = -\frac{3}{2} = -1.5\]
We must verify that our solution \(x = -1.5\) lies within the domain. Since \(-1.5
eq 1\) and \(-1.5
eq -1\), the solution is valid. We can substitute \(x = -1.5\) back into the original equation to confirm:
LHS: \(\frac{3}{-1.5-1} - \frac{1}{(-1.5)^2-1} = \frac{3}{-2.5} - \frac{1}{2.25-1} = -1.2 - \frac{1}{1.25} = -1.2 - 0.8 = -2\)
RHS: \(\frac{5}{-1.5-1} = \frac{5}{-2.5} = -2\)
Both sides are equal, confirming the result. Thus, the solution is \(\boxed{-1.5}\).
Solving rational equations is vital in various fields, such as engineering and physics, where we deal with rates and ratios. For instance, if you are calculating the time it takes for two different pumps to fill a reservoir, you will often encounter rational expressions. By setting up an equation involving these rates and solving for the unknown, you can determine exactly when the job will be completed or when certain conditions are met, ensuring efficient resource management and planning.