The width of a rectangle is 3 inches less than the length. the area is 50 square inches. find the length and width. round to the nearest tenth if necessary

the length is 8.7 inches. The width is 5.7 inches

Step-by-step explanation:

Let L represent the length of the rectangle.

Let W represent the width of the rectangle.

The width of a rectangle is 3 inches less than the length. This means that

L = W + 3

The area of a rectangle is expressed as L×W

The area is 50 square inches. This means that

LW = 50 - - - - - - - - - - 1

Substituting L = W + 3 into equation 1, it becomes

(W + 3)W = 50

W^2 + 3W - 50 = 0

We would apply the general formula for quadratic equations,

x = [ - b ± √(b^2 - 4ac)]/2a

a = 1

b = 3

c = - 50

Therefore,

w = [- 3 ± √(3^2 - 4×1×-50)]/2×1

w = [- 3 ± √(9 + 200)]/2

w = (- 3 ± √209)/2

w = (-3+14.4568)/2 or (-3-14.4568)/2

w = 5.7 or w = -8.7

Since the width cannot be negative, the width would be 5.7 inches.

Substituting w = 5.7 into L = W + 3,

L = 5.7 + 3 = 8.7 inches

answer: all the values are within because it has a pattern that repeats from -6 to 4.

good luck!

the first one, all those points on the line just make a strait line.