You have a system of equations:

1. \( X + Y = 166 \)
2. \( X \times Y = 5520 \)

To solve for \(X\) and \(Y\), we can express \(Y\) in terms of \(X\) using the first equation:

\[
Y = 166 - X
\]

Now, substitute \(Y\) into the second equation:

\[
X \times (166 - X) = 5520
\]

Expanding this, we get:

\[
166X - X^2 = 5520
\]

Rearranging gives us a standard quadratic equation:

\[
X^2 - 166X + 5520 = 0
\]

Next, we can use the quadratic formula to find \(X\):

\[
X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \(a = 1\), \(b = -166\), and \(c = 5520\):

1. Calculate the discriminant:
   \[
   b^2 - 4ac = (-166)^2 - 4 \times 1 \times 5520 = 27556 - 22080 = 5476
   \]

2. Now apply the quadratic formula:
   \[
   X = \frac{166 \pm \sqrt{5476}}{2}
   \]
   \[
   \sqrt{5476} = 74 \quad \text{(since \(74^2 = 5476\))}
   \]
   \[
   X = \frac{166 \pm 74}{2}
   \]

This gives us two solutions:

\[
X_1 = \frac{240}{2} = 120, \quad X_2 = \frac{92}{2} = 46
\]

Now, substituting back to find \(Y\):

1. If \(X = 120\):
   \[
   Y = 166 - 120 = 46
   \]

2. If \(X = 46\):
   \[
   Y = 166 - 46 = 120
   \]

Thus, the solutions are:

\[
(X, Y) = (120, 46) \quad \text{or} \quad (46, 120)
\]

Would you like further details on any part of the solution? Here are some related questions you might consider:

1. How can we verify the solutions?
2. What happens if we change the values in the equations?
3. How can we graph these equations?
4. What is the significance of the discriminant in this context?
5. How do systems of equations apply in real-world scenarios?

**Tip:** Always check your solutions by substituting them back into the original equations!

Discover how to solve the system of equations X + Y = 166 and X * Y = 5520 through a detailed step-by-step guide. This problem explores algebraic concepts, including quadratic equations and the quadratic formula. Ideal for students in Grades 9-12.

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