https://file.aiquickdraw.com/mathbot/file/172499551139968ymlp74.jpg

The image contains two sets of equations to solve. Let's solve each set step by step.

### Problem 1: Solve the system of equations

\[
\begin{aligned}
1.\quad & x^2 + y^2 = 29 \\
2.\quad & x - y = 3
\end{aligned}
\]

#### Step 1: Express \( x \) from the second equation

From the second equation:

\[
x = y + 3
\]

#### Step 2: Substitute \( x = y + 3 \) into the first equation

Substituting into the first equation:

\[
(y + 3)^2 + y^2 = 29
\]

Expand and simplify:

\[
y^2 + 6y + 9 + y^2 = 29
\]

Combine like terms:

\[
2y^2 + 6y + 9 = 29
\]

Subtract 29 from both sides:

\[
2y^2 + 6y - 20 = 0
\]

Divide the entire equation by 2 to simplify:

\[
y^2 + 3y - 10 = 0
\]

#### Step 3: Solve the quadratic equation

The quadratic equation can be solved using the quadratic formula:

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

For \( a = 1 \), \( b = 3 \), and \( c = -10 \):

\[
y = \frac{-3 \pm \sqrt{3^2 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2} = \frac{-3 \pm 7}{2}
\]

So:

\[
y = \frac{4}{2} = 2 \quad \text{or} \quad y = \frac{-10}{2} = -5
\]

#### Step 4: Find \( x \)

Using \( x = y + 3 \):

- If \( y = 2 \), then \( x = 2 + 3 = 5 \).
- If \( y = -5 \), then \( x = -5 + 3 = -2 \).

#### Final Solutions

The solutions are:

\[
(x, y) = (5, 2) \quad \text{or} \quad (-2, -5)
\]

### Problem 2: Solve the system of equations

\[
\begin{aligned}
1.\quad & \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} = \frac{5}{2} \\
2.\quad & x + y = 10
\end{aligned}
\]

#### Step 1: Set up substitutions

Let:

\[
\sqrt{\frac{x}{y}} = a \quad \text{and} \quad \sqrt{\frac{y}{x}} = \frac{1}{a}
\]

Thus, the first equation becomes:

\[
a + \frac{1}{a} = \frac{5}{2}
\]

#### Step 2: Multiply both sides by \( a \)

This gives:

\[
a^2 + 1 = \frac{5a}{2}
\]

Multiply by 2 to clear the fraction:

\[
2a^2 - 5a + 2 = 0
\]

#### Step 3: Solve the quadratic equation

Using the quadratic formula:

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

For \( a = 2 \), \( b = -5 \), and \( c = 2 \):

\[
a = \frac{5 \pm \sqrt{(-5)^2 - 16}}{4} = \frac{5 \pm \sqrt{9}}{4} = \frac{5 \pm 3}{4}
\]

So:

\[
a = 2 \quad \text{or} \quad a = \frac{1}{2}
\]

#### Step 4: Relate to \( x \) and \( y \)

For \( a = 2 \):

\[
\sqrt{\frac{x}{y}} = 2 \quad \Rightarrow \quad \frac{x}{y} = 4 \quad \Rightarrow \quad x = 4y
\]

For \( a = \frac{1}{2} \):

\[
\sqrt{\frac{x}{y}} = \frac{1}{2} \quad \Rightarrow \quad \frac{x}{y} = \frac{1}{4} \quad \Rightarrow \quad x = \frac{y}{4}
\]

#### Step 5: Use \( x + y = 10 \)

For \( x = 4y \):

\[
4y + y = 10 \quad \Rightarrow \quad 5y = 10 \quad \Rightarrow \quad y = 2, \, x = 8
\]

For \( x = \frac{y}{4} \):

\[
\frac{y}{4} + y = 10 \quad \Rightarrow \quad \frac{5y}{4} = 10 \quad \Rightarrow \quad y = 8, \, x = 2
\]

#### Final Solutions

The solutions are:

\[
(x, y) = (8, 2) \quad \text{or} \quad (2, 8)
\]

---

Would you like more details or have any other questions?

Here are some related questions:

1. How does solving quadratic equations help in systems of equations?
2. What is the significance of substituting in solving equations with radicals?
3. Can systems with radicals have extraneous solutions?
4. What is the importance of checking all possible solutions in non-linear equations?
5. How can graphical methods be used to solve systems of equations?

**Tip:** When dealing with radicals in equations, always check if the solutions satisfy the original equation to avoid including extraneous solutions.

Learn how to solve complex systems of equations involving quadratics and radicals step-by-step. This problem covers key concepts including the quadratic formula and solving equations with radicals. Suitable for high school students (Grades 10-12).

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation