the perimeter of a rectangle is 222 inches. the length exceeds the width by 47 inches. what is the length and width

width -- x

length -- x+47

solve
2(x + x+47) = 222

P=2l +2w

222 = 2(w+47) + 2w
222= 2w +94 +2w
222=4w+94
222-94=4w+94-94
128=4w
32=w so length = w+47
length = 32+47=79

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

Let's assume the width of the rectangle is x inches. According to the problem, the length exceeds the width by 47 inches, so the length would be (x + 47) inches.

The formula for calculating the perimeter of a rectangle is P = 2(length + width). In this case, the perimeter is given as 222 inches. Plugging in the values, we have:

222 = 2(x + 47 + x)

Simplifying this equation, we get:

222 = 2(2x + 47)
222 = 4x + 94
222 - 94 = 4x
128 = 4x
x = 128/4
x = 32

So, the width of the rectangle is 32 inches. To find the length, we substitute this value back into our expression for length:

Length = x + 47
Length = 32 + 47
Length = 79

Therefore, the length of the rectangle is 79 inches and the width is 32 inches.