Question

solve for all values of x:
$1 + \frac{x}{x + 1} = \frac{1}{x^2 + x}$

Explanation:

Step1: Simplify the equation

First, factor the denominator on the right side: \(x^2 + x = x(x + 1)\). The original equation is \(1+\frac{x}{x + 1}=\frac{1}{x(x + 1)}\). Multiply each term by \(x(x + 1)\) (the least common denominator) to eliminate the fractions. We get:
\(x(x + 1)\times1+x(x + 1)\times\frac{x}{x + 1}=x(x + 1)\times\frac{1}{x(x + 1)}\)
Simplify each term:
\(x(x + 1)+x\times x = 1\)
Expand \(x(x + 1)\): \(x^2+x+x^2 = 1\)
Combine like terms: \(2x^2+x - 1 = 0\)

Step2: Solve the quadratic equation

We have a quadratic equation \(2x^2+x - 1 = 0\). Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 1\), \(c=-1\).
First, calculate the discriminant \(\Delta=b^2 - 4ac=(1)^2-4\times2\times(-1)=1 + 8 = 9\)
Then, \(x=\frac{-1\pm\sqrt{9}}{2\times2}=\frac{-1\pm3}{4}\)
We get two solutions:
\(x_1=\frac{-1 + 3}{4}=\frac{2}{4}=\frac{1}{2}\)
\(x_2=\frac{-1-3}{4}=\frac{-4}{4}=-1\)

Step3: Check for extraneous solutions

Now, check if these solutions make the original denominators zero.
For \(x=-1\), the denominators \(x + 1\) and \(x(x + 1)\) are zero, so \(x=-1\) is extraneous.
For \(x=\frac{1}{2}\), the denominators \(x + 1=\frac{1}{2}+1=\frac{3}{2}
eq0\) and \(x(x + 1)=\frac{1}{2}\times\frac{3}{2}=\frac{3}{4}
eq0\), so \(x=\frac{1}{2}\) is a valid solution.

Answer:

\(x = \frac{1}{2}\)