Solve the System of Equations: Square Roots and xy=9

To solve this problem, we will follow these steps:

• Step 1: Identify the equations and express one variable in terms of the other.
• Step 2: Substitute into the other equation and simplify.
• Step 3: Perform calculations to solve for the variable.
• Step 4: Use the solution to find the second variable.

Let's work through the solution together:

Step 1: Given $xy = 9$, express $y$ as $\frac{9}{x}$.

Step 2: Substitute into the first equation:

$\sqrt{x} + \sqrt{\frac{9}{x}} = \sqrt{\sqrt{61} + 6}$.

Step 3: Simplify this equation. Let $a = \sqrt{x}$ and $b = \sqrt{y}$.

Then, $a + b = \sqrt{\sqrt{61} + 6}$ and $ab = \sqrt{9} = 3$.

Squaring both sides of the linear equation:

$(a + b)^2 = \sqrt{61} + 6$.

$a^2 + 2ab + b^2 = \sqrt{61} + 6$.

Using $ab = 3$, we get $2ab = 6$.

This leads to $a^2 + b^2 = \sqrt{61}$.

Replacing $a = \sqrt{x}$ and $b = \sqrt{y}$:

Let $a^2 = x$ and $b^2 = y$ and use the identity $(a - b)^2 = a^2 + b^2 - 2ab = \sqrt{61} - 6$.

So, $a - b = \sqrt{\sqrt{61} - 6}$.

Now let $S = a + b$ and $P = ab$ from previous steps.

From $S = \sqrt{\sqrt{61} + 6}$ and $P = 3$, solve: $t^2 - St + P = 0$.

This quadratic in $t$ gives solutions $t = \frac{S \pm \sqrt{S^2 - 4P}}{2}$.

The quadratic roots are $a = \frac{\sqrt{61} + 6}{2} \pm \frac{5}{2}$ and $b = \frac{\sqrt{61} + 6}{2} \mp \frac{5}{2}$.

Thus, $x = a^2 = (\frac{\sqrt{61}}{2} + 2.5)^2$ or $(\frac{\sqrt{61}}{2} - 2.5)^2$.

Similarly for $y$.

Therefore, the solutions are:

$x = \frac{\sqrt{61}}{2} - 2.5$, $y = \frac{\sqrt{61}}{2} + 2.5$

or

$x = \frac{\sqrt{61}}{2} + 2.5$, $y = \frac{\sqrt{61}}{2} - 2.5$.