We have an equation containing the expression: \(\frac{x+1}{x}\). Let's perform a substitution:
Let: \(\frac{x+1}{x} = a\)
The original equation transforms into:
\[a^2 - 3a + 2 = 0\]Solving this quadratic equation yields:
\[a = 1 \quad \text{or} \quad a = 2\] Analysis of Each Value of \(a\)
- Case 1: \(a = 2 \Rightarrow \frac{x+1}{x} = 2\)
Multiplying both sides by \(x\) gives: \(x + 1 = 2x \Rightarrow x = 1\)
This is a valid real solution ✅ - Case 2: \(a = 1 \Rightarrow \frac{x+1}{x} = 1\)
Multiplying both sides by \(x\) gives: \(x + 1 = x \Rightarrow 1 = 0\)
❌ This is a contradiction, hence not a valid solution.
Conclusion
Therefore, the sole real solution is: \(x = 1\)
Total number of real solutions = 1