A rectangle has a length of 5a+2b and a width of 4a−3b. Find the equations for the perimeter and the area of the rectangle

To find the equations for the perimeter and the area of the rectangle, let's use the given dimensions:

- **Length (\( L \))**: \( 5a + 2b \)
- **Width (\( W \))**: \( 4a - 3b \)

### Perimeter of the Rectangle
The perimeter \( P \) of a rectangle is given by:
\[
P = 2L + 2W
\]
Substituting the values for \( L \) and \( W \):
\[
P = 2(5a + 2b) + 2(4a - 3b)
\]
Distribute the 2:
\[
P = 10a + 4b + 8a - 6b
\]
Now, combine like terms:
\[
P = (10a + 8a) + (4b - 6b) = 18a - 2b
\]
So, the equation for the perimeter is:
\[
\boxed{P = 18a - 2b}
\]

### Area of the Rectangle
The area \( A \) of a rectangle is given by:
\[
A = L \times W
\]
Substituting the values for \( L \) and \( W \):
\[
A = (5a + 2b)(4a - 3b)
\]
Apply the distributive property (FOIL method):
\[
A = (5a)(4a) + (5a)(-3b) + (2b)(4a) + (2b)(-3b)
\]
Simplify each term:
\[
A = 20a^2 - 15ab + 8ab - 6b^2
\]
Now, combine like terms:
\[
A = 20a^2 - 7ab - 6b^2
\]
So, the equation for the area is:
\[
\boxed{A = 20a^2 - 7ab - 6b^2}
\]

Would you like more details or have any other questions?

Here are some related questions:
1. What is the perimeter when \( a = 2 \) and \( b = 1 \)?
2. What is the area when \( a = 1 \) and \( b = 2 \)?
3. How would the perimeter change if the width were \( 4a + 3b \) instead of \( 4a - 3b \)?
4. What happens to the area if both \( a \) and \( b \) are doubled?
5. Can you find the length and width if the perimeter is given as \( 36a + 2b \)?

**Tip:** When multiplying binomials, remember to use the distributive property carefully to ensure all terms are accounted for.

Learn how to find the perimeter and area of a rectangle with dimensions 5a+2b (length) and 4a−3b (width). This problem involves algebraic manipulation and application of perimeter and area formulas. Suitable for Grades 7-9 students learning geometric concepts.

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