\left(3+x\sqrt{5}\right)\left(\sqrt{5}-y\right)=-13+5\sqrt{5}

Let's solve the equation step by step.

We are given the equation:

\[
(3 + x\sqrt{5})(\sqrt{5} - y) = -13 + 5\sqrt{5}
\]

### Step 1: Expand the left-hand side
We use the distributive property (FOIL method) to expand the left-hand side:

\[
(3 + x\sqrt{5})(\sqrt{5} - y) = 3(\sqrt{5}) + 3(-y) + x\sqrt{5}(\sqrt{5}) + x\sqrt{5}(-y)
\]

Simplify each term:

\[
= 3\sqrt{5} - 3y + x(\sqrt{5})(\sqrt{5}) - xy\sqrt{5}
\]

Since \(\sqrt{5} \times \sqrt{5} = 5\), we get:

\[
= 3\sqrt{5} - 3y + 5x - xy\sqrt{5}
\]

Thus, the expanded form is:

\[
= (3\sqrt{5} - xy\sqrt{5}) + (5x - 3y)
\]

### Step 2: Group terms with \(\sqrt{5}\)
We can group the terms with and without \(\sqrt{5}\):

\[
= (3 - xy)\sqrt{5} + (5x - 3y)
\]

### Step 3: Compare both sides
The equation now is:

\[
(3 - xy)\sqrt{5} + (5x - 3y) = -13 + 5\sqrt{5}
\]

Now, equate the terms with \(\sqrt{5}\) and the constant terms separately.

1. For the \(\sqrt{5}\) terms:
\[
3 - xy = 5
\]

2. For the constant terms:
\[
5x - 3y = -13
\]

### Step 4: Solve the system of equations
We now have the following system of equations:

1. \(3 - xy = 5\), which simplifies to:
\[
xy = -2
\]

2. \(5x - 3y = -13\)

Now, solve this system of equations.

### Step 5: Solve the system
From the first equation, \(xy = -2\), express \(y\) in terms of \(x\):

\[
y = \frac{-2}{x}
\]

Substitute this into the second equation:

\[
5x - 3\left(\frac{-2}{x}\right) = -13
\]

Simplify:

\[
5x + \frac{6}{x} = -13
\]

### Step 6: Multiply through by \(x\) to eliminate the fraction
Multiply the entire equation by \(x\) (assuming \(x \neq 0\)):

\[
5x^2 + 6 = -13x
\]

Rearrange the equation:

\[
5x^2 + 13x + 6 = 0
\]

### Step 7: Solve the quadratic equation
We now solve the quadratic equation \(5x^2 + 13x + 6 = 0\). Use the quadratic formula:

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

where \(a = 5\), \(b = 13\), and \(c = 6\).

\[
x = \frac{-13 \pm \sqrt{13^2 - 4(5)(6)}}{2(5)}
\]
\[
x = \frac{-13 \pm \sqrt{169 - 120}}{10}
\]
\[
x = \frac{-13 \pm \sqrt{49}}{10}
\]
\[
x = \frac{-13 \pm 7}{10}
\]

So, we have two solutions for \(x\):

\[
x = \frac{-13 + 7}{10} = \frac{-6}{10} = -\frac{3}{5}
\]
\[
x = \frac{-13 - 7}{10} = \frac{-20}{10} = -2
\]

### Step 8: Solve for \(y\)
Now substitute each value of \(x\) back into the equation \(xy = -2\) to solve for \(y\).

1. When \(x = -\frac{3}{5}\):
\[
-\frac{3}{5}y = -2 \quad \Rightarrow \quad y = \frac{-2}{-\frac{3}{5}} = \frac{10}{3}
\]

2. When \(x = -2\):
\[
-2y = -2 \quad \Rightarrow \quad y = 1
\]

### Step 9: Final solutions
The solutions are:

\[
(x, y) = \left(-\frac{3}{5}, \frac{10}{3}\right) \quad \text{or} \quad (x, y) = (-2, 1)
\]

Would you like more details or have any questions about this process?

---

Here are 5 related questions:
1. How can you verify if both solutions satisfy the original equation?
2. What happens if we change the constants in the equation?
3. How does the distributive property work for complex numbers?
4. Can you graph the solution pairs for visual verification?
5. How would the solution process change if the square root term were different?

**Tip:** When solving quadratic equations, always check if factoring is possible before applying the quadratic formula for efficiency.

Explore how to solve the system of equations involving square roots and quadratic expressions with a detailed step-by-step solution. This problem demonstrates key algebraic techniques such as the distributive property, the quadratic formula, and solving systems of equations. Suitable for students in Grades 9-11.

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