https://file.aiquickdraw.com/mathbot/file/17280397281751fus8336.jpg

The diagram shows a square with four labeled sides:

- The top side is \( 6(x + 1) \).
- The bottom side is \( 3x + 21 \).
- The left side is \( \frac{y}{3} + 16 \).
- The right side is \( 20 \).

Since it is a square, all sides must be equal in length. This gives us two equations to solve for \( x \) and \( y \):

1. \( 6(x + 1) = 3x + 21 \) (equating the top and bottom sides)
2. \( \frac{y}{3} + 16 = 20 \) (equating the left and right sides)

### Solving for \( x \):
\[
6(x + 1) = 3x + 21
\]
Expanding the left side:
\[
6x + 6 = 3x + 21
\]
Now subtract \( 3x \) from both sides:
\[
3x + 6 = 21
\]
Subtract 6 from both sides:
\[
3x = 15
\]
Now divide by 3:
\[
x = 5
\]

### Solving for \( y \):
\[
\frac{y}{3} + 16 = 20
\]
Subtract 16 from both sides:
\[
\frac{y}{3} = 4
\]
Now multiply by 3:
\[
y = 12
\]

Thus, the values of \( x \) and \( y \) are:
- \( x = 5 \)
- \( y = 12 \)

Would you like a detailed explanation of any step or have further questions?

Here are five related questions to deepen understanding:
1. What happens if the sides were rectangles instead of squares?
2. How would you solve this if you had inequalities instead of equations?
3. Can you form another equation involving \( x \) or \( y \) to check consistency?
4. What is the perimeter of the square given these values of \( x \) and \( y \)?
5. How would the method change if the problem involved a cube?

**Tip:** When solving equations, always remember to check your solutions by substituting them back into the original equations to verify.

Work out the values of x and y in the diagram of a square where the side lengths are algebraic expressions: top side = 6(x + 1), bottom side = 3x + 21, left side = y/3 + 16, and right side = 20.

Learn how to solve for x and y in a square where the sides are expressed algebraically: top side = 6(x + 1), bottom side = 3x + 21, left side = y/3 + 16, and right side = 20. Suitable for students in Grades 8-10.

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