Question

solve the rational equation
\\(\frac{x}{x + 1} + \frac{5}{x - 1} = 1\\)

Explanation:

Step1: Find the common denominator

The common denominator of \(x + 1\) and \(x - 1\) is \((x + 1)(x - 1)\). Multiply each term by \((x + 1)(x - 1)\) to eliminate the fractions:

$$ \frac{x}{x + 1} \cdot (x + 1)(x - 1) + \frac{5}{x - 1} \cdot (x + 1)(x - 1) = 1 \cdot (x + 1)(x - 1) $$

Step2: Simplify each term

Simplify the left - hand side and the right - hand side:
For the first term: \(\frac{x}{x + 1}\cdot(x + 1)(x - 1)=x(x - 1)=x^{2}-x\)
For the second term: \(\frac{5}{x - 1}\cdot(x + 1)(x - 1)=5(x + 1)=5x + 5\)
For the right - hand side: \(1\cdot(x + 1)(x - 1)=x^{2}-1\)
So the equation becomes:

$$ x^{2}-x + 5x + 5=x^{2}-1 $$

Step3: Combine like terms

Combine the \(x\) terms on the left - hand side:
\(x^{2}+4x + 5=x^{2}-1\)

Step4: Solve for \(x\)

Subtract \(x^{2}\) from both sides of the equation:
\(4x+5=-1\)
Subtract 5 from both sides:
\(4x=-1 - 5=-6\)
Divide both sides by 4:
\(x =-\frac{6}{4}=-\frac{3}{2}\)
We need to check if this solution makes the original denominators zero. For \(x=-\frac{3}{2}\), \(x + 1=-\frac{3}{2}+1=-\frac{1}{2}
eq0\) and \(x - 1=-\frac{3}{2}-1=-\frac{5}{2}
eq0\). So \(x =-\frac{3}{2}\) is a valid solution.

Answer:

\(x =-\frac{3}{2}\)