To solve the problems:

Problem 22: Write an equation to solve for $x$.

The area of a rectangle is given as:

$\text{Area} = \text{length} \times \text{width}$

Here, the length is $20 + 5x$ and the width is $10$. The area is given as $31x$.

So, the equation becomes:

$(20 + 5x) \times 10 = 31x$

Simplify the equation:

$200 + 50x = 31x$

Rearranging:

$50x - 31x = -200$

$19x = -200$

$x = -\frac{200}{19}$

So, $x = -\frac{200}{19}$.

Problem 23: Write the expression to find the perimeter of the rectangle in simplest form.

The perimeter of a rectangle is given by:

$\text{Perimeter} = 2(\text{length} + \text{width})$

Substituting the length $20 + 5x$ and the width $10$:

$\text{Perimeter} = 2((20 + 5x) + 10)$

$\text{Perimeter} = 2(30 + 5x)$

$\text{Perimeter} = 60 + 10x$

So, the expression for the perimeter in simplest form is $60 + 10x$.

Would you like further details or explanations on any of these steps?

Related Questions:

1. What would be the length of the rectangle when $x = -\frac{200}{19}$?
2. How does changing $x$ affect the area and perimeter of the rectangle?
3. What happens if the area of the rectangle were different, say $50x$ instead of $31x$?
4. Can the value of $x$ ever be positive given these conditions? Why or why not?
5. How would the equation change if the width of the rectangle was $15$ instead of $10$?

Tip:

Always double-check your work by substituting your solution back into the equation to verify if it satisfies the given conditions.