The given system of equations is:

$x^2 + y^2 - 4x + 3 = 0$

$y^2 + x - 1 = 0$

Solution Approach:

1. Express $y^2$ in terms of $x$ from the second equation: $y^2 = 1 - x$

2. Substitute $y^2 = 1 - x$ into the first equation: $x^2 + (1 - x) - 4x + 3 = 0$

3. Simplify the equation: $x^2 - x + 1 - 4x + 3 = 0$ $x^2 - 5x + 4 = 0$

4. Factorize the quadratic equation: $(x - 4)(x - 1) = 0$

   So, the solutions for $x$ are: $x = 4 \quad \text{or} \quad x = 1$

5. Find corresponding $y$ values:

   • For $x = 4$:

     $y^2 = 1 - 4 = -3$ (No real solution since $y^2$ cannot be negative)

   • For $x = 1$:

     $y^2 = 1 - 1 = 0$

     So, $y = 0$.

Conclusion:

The only real solution is:

$(1, 0)$

Would you like a detailed explanation of any step? Here are some related questions you might find useful:

1. How do you determine the real solution set of a system of equations?
2. What methods can be used to solve a system of nonlinear equations?
3. How do you check if a quadratic equation has real solutions?
4. What is the graphical interpretation of this solution?
5. Can a system of equations have no real solutions? How do you identify such cases?

Tip: When solving nonlinear systems, substitution is a powerful method, especially when one equation can be solved explicitly for a variable.