The area of the conference table in mr. nathan’s office must be no more than 175 ft2. if the length of the table is 18 ft more than the width, x, which interval can be the possible widths?

LW < = A (thats less then or equal to)
W = x
L = x + 18
A = 175

x(x + 18) < = 175
x^2 + 18x - 175 < = 0
(x + 25)(x - 7) < = 0

x + 25 < = 0
x < = -25extraneous solution...not ur answer

x - 7 < = 0
x < = 7 <==

so anything less then or equal to 7 can be a possible width