Find all ordered pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 62$.

We are given the system of equations:

\[
x + y = 10
\]
\[
x^2 + y^2 = 62
\]

We need to find all ordered pairs $(x, y)$ that satisfy both equations.

### Step 1: Express $y$ in terms of $x$
From the first equation, we can express $y$ in terms of $x$:

\[
y = 10 - x
\]

### Step 2: Substitute $y = 10 - x$ into the second equation
Substitute $y = 10 - x$ into the second equation:

\[
x^2 + (10 - x)^2 = 62
\]

### Step 3: Expand $(10 - x)^2$
Expand the square:

\[
(10 - x)^2 = 100 - 20x + x^2
\]

Substituting this into the equation gives:

\[
x^2 + 100 - 20x + x^2 = 62
\]

### Step 4: Simplify and solve the quadratic equation
Combine like terms:

\[
2x^2 - 20x + 100 = 62
\]

Subtract 62 from both sides:

\[
2x^2 - 20x + 38 = 0
\]

### Step 5: Divide by 2
To simplify, divide the entire equation by 2:

\[
x^2 - 10x + 19 = 0
\]

### Step 6: Solve the quadratic equation
Solve the quadratic equation using the quadratic formula:

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

Here, $a = 1$, $b = -10$, and $c = 19$:

\[
x = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(19)}}{2(1)}
\]

\[
x = \frac{10 \pm \sqrt{100 - 76}}{2}
\]

\[
x = \frac{10 \pm \sqrt{24}}{2}
\]

\[
x = \frac{10 \pm 2\sqrt{6}}{2}
\]

\[
x = 5 \pm \sqrt{6}
\]

### Step 7: Find the corresponding $y$ values
Using $y = 10 - x$:

- If $x = 5 + \sqrt{6}$:

\[
y = 10 - (5 + \sqrt{6}) = 5 - \sqrt{6}
\]

- If $x = 5 - \sqrt{6}$:

\[
y = 10 - (5 - \sqrt{6}) = 5 + \sqrt{6}
\]

### Final Answer
The ordered pairs $(x, y)$ that satisfy the system are:

\[
(5 + \sqrt{6}, 5 - \sqrt{6}) \quad \text{and} \quad (5 - \sqrt{6}, 5 + \sqrt{6})
\]

These pairs satisfy both given equations.

Would you like further clarification on any step or have other questions?

---
Here are 8 related questions for further exploration:

1. How would you verify that the solutions satisfy both equations?
2. Can you derive the equations of circles and lines that relate to the problem?
3. What happens if the system $x + y = a$ and $x^2 + y^2 = b$ has no solutions?
4. How can completing the square help in solving quadratic equations?
5. What are the geometric interpretations of the given equations?
6. How would you find the roots of $x^2 - 10x + 19 = 0$ using factoring?
7. How does the quadratic formula work in different scenarios, like when the discriminant is zero?
8. Can you generalize the solution method to similar problems with different constants?

**Tip:** When solving systems involving both linear and quadratic equations, substitution is often the most effective strategy.

Learn how to solve the system of equations x + y = 10 and x^2 + y^2 = 62 with a step-by-step solution. This problem involves quadratic equations and the substitution method, suitable for students in Grades 10-12.

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