SOLUTION: the area of the rectangle shown is 40in^2. find the length and width. Length of one side is x+7, and the width of the other side is 2x+3. Thank you for taking the time to read this
Algebra ->
Rectangles
-> SOLUTION: the area of the rectangle shown is 40in^2. find the length and width. Length of one side is x+7, and the width of the other side is 2x+3. Thank you for taking the time to read this
Log On
Question 528814: the area of the rectangle shown is 40in^2. find the length and width. Length of one side is x+7, and the width of the other side is 2x+3. Thank you for taking the time to read this. Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! the area of the rectangle shown is 40in^2. find the length and width. Length of one side is x+7, and the width of the other side is 2x+3.
(x+7)(2x+3) = 40
FOIL the left to get:
2x^2+3x+14x+21 = 40
2x^2+17x+21 = 40
2x^2+17x-19 = 0
2x^2-2x+19x-19 = 0
(2x^2-2x)+(19x-19) = 0
2x(x-1)+19(x-1) = 0
(x-1)(2x+19) = 0
x = {-19, 1}
throw out the negative solution (extraneous) leaving:
x = 1
.
Length:
x+7 = 1+7 = 8 inches
.
Width:
2x+3 = 2(1)+3 = 2+3 = 5 inches
