The problem in the image is asking for the number of real solutions $(x, y)$ that satisfy the following system of equations:

$3x^2 + 2y^2 = 1$ $2x^2 + y^2 = 3$

Step 1: Subtract the equations

First, let's subtract the second equation from the first:

$(3x^2 + 2y^2) - (2x^2 + y^2) = 1 - 3$

Simplifying:

$x^2 + y^2 = -2$

Step 2: Analyze the result

The equation $x^2 + y^2 = -2$ has no real solutions. This is because the sum of two squares, $x^2 + y^2$, is always non-negative for real numbers $x$ and $y$, and it cannot be negative. Therefore, the system of equations has no real solutions.

Conclusion:

The number of real solutions is 0.

However, "0" is not listed as one of the options, so the correct assumption here is that the answer should be none of the provided options (since real solutions don't exist).

Would you like any more clarification, or should we dive deeper into any specific part?

Related questions:

1. What does the sum of squares mean in terms of real numbers?
2. How do you solve systems of quadratic equations?
3. Why is it impossible for $x^2 + y^2 = -2$ to have real solutions?
4. Can quadratic systems have no real solutions but still have complex solutions?
5. How do you analyze and graph the equations to check for intersections?

Tip:

When solving systems of equations, subtracting or adding the equations can often help in reducing complexity and revealing hidden insights.