https://file.aiquickdraw.com/mathbot/file/17261597052580f4qzavd.jpg

Let's solve the equations for \(x\) and \(y\) in each figure.

### 1. **Square:**
For a square, all four sides are equal. So, the expressions representing the sides should be equal.

- The top and bottom sides are given as \(2x + 1\) and \(3x - 4\), respectively.
- The left and right sides are given as \(y - 5\) and \(x + 6\), respectively.

**Equating the sides:**

\[
2x + 1 = 3x - 4
\]

Solve for \(x\):

\[
2x - 3x = -4 - 1
\]
\[
-x = -5
\]
\[
x = 5
\]

Next, equate the other pair:

\[
y - 5 = x + 6
\]

Substitute \(x = 5\):

\[
y - 5 = 5 + 6
\]
\[
y - 5 = 11
\]
\[
y = 16
\]

So, for the square:
\[
x = 5, \quad y = 16
\]

### 2. **Kite:**
For a kite, adjacent sides are equal.

- The top left and top right sides are given as \(3y + 2\) and \(2x + 3\), respectively.
- The bottom left and bottom right sides are given as \(y + 4\) and \(5x - 2\), respectively.

**Equating the pairs of adjacent sides:**

For the top sides:

\[
3y + 2 = 2x + 3
\]

For the bottom sides:

\[
y + 4 = 5x - 2
\]

Now, solve these simultaneously:

1. Solve the first equation for \(x\):

\[
3y + 2 = 2x + 3 \Rightarrow 2x = 3y - 1 \Rightarrow x = \frac{3y - 1}{2}
\]

2. Substitute \(x = \frac{3y - 1}{2}\) into the second equation:

\[
y + 4 = 5\left(\frac{3y - 1}{2}\right) - 2
\]
\[
y + 4 = \frac{15y - 5}{2} - 2
\]
\[
2(y + 4) = 15y - 5 - 4
\]
\[
2y + 8 = 15y - 9
\]
\[
8 + 9 = 15y - 2y
\]
\[
17 = 13y \Rightarrow y = \frac{17}{13}
\]

Now substitute \(y = \frac{17}{13}\) back into \(x = \frac{3y - 1}{2}\):

\[
x = \frac{3\left(\frac{17}{13}\right) - 1}{2} = \frac{\frac{51}{13} - 1}{2} = \frac{\frac{51}{13} - \frac{13}{13}}{2} = \frac{\frac{38}{13}}{2} = \frac{38}{26} = \frac{19}{13}
\]

So, for the kite:
\[
x = \frac{19}{13}, \quad y = \frac{17}{13}
\]

### Summary:
- **Square:** \(x = 5\), \(y = 16\)
- **Kite:** \(x = \frac{19}{13}\), \(y = \frac{17}{13}\)

Would you like further details or have any questions?

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**Related Questions:**
1. How would the equations change if the square had a different set of expressions for its sides?
2. What are the properties of a kite that lead to these specific equations?
3. How could we check the correctness of the calculated \(x\) and \(y\) values in these figures?
4. Can we generalize a method to solve for \(x\) and \(y\) in other quadrilaterals?
5. What would the equations look like if one diagonal of the kite was also given?

**Tip:** When solving simultaneous equations, it's often helpful to first express one variable in terms of the other to simplify substitution.

Find the values of x and y for a square and a kite, based on the provided algebraic expressions for the sides.

In this problem, we solve for x and y using algebraic equations for the sides of a square and a kite. The properties of squares (all sides equal) and kites (adjacent sides equal) help us derive the equations and find the values of x and y. Suitable for Grade 9-10 students.

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