The perimeter of a rectangle is 174 inches. The length exceeds the width by 11 inches. find the length and the width.

To find the length and width of the rectangle, we can set up a system of equations based on the given information.

Let's assume:
- Length of the rectangle = L inches
- Width of the rectangle = W inches

Given that:
- The perimeter of a rectangle is 174 inches, we can write the equation: 2L + 2W = 174.

Furthermore, it is given that:
- The length exceeds the width by 11 inches, so we can write the equation: L = W + 11.

We now have a system of two equations:
1) 2L + 2W = 174
2) L = W + 11

To solve this system, we can substitute equation (2) into equation (1) to eliminate L:

2(W + 11) + 2W = 174
Simplifying, 2W + 22 + 2W = 174
Combine like terms, 4W + 22 = 174
Subtract 22 from both sides, 4W = 152
Divide both sides by 4, W = 38

We now know that the width of the rectangle is 38 inches.

Using the value of W, we can substitute it back into equation (2) to find the length:

L = 38 + 11
L = 49

Therefore, the length of the rectangle is 49 inches.

In conclusion, the length of the rectangle is 49 inches and the width is 38 inches.

W=38 but then how do you find length?

The length exceeds the width by 11 inches.

P = 2L + 2W

174 = W + 11 + W + 11 + 2W
174 = 4W + 22
152 = 4W
152/4 = W
? = width