SOLUTION: Choose a possible dimensions of a rectangle with a width= 2x-1 and length= 3x+2 so that its area is greater than 153cm² a. W= 13 L= 15 b. W= 11 L= 20 c. W= 11 L= 21 d. W= 13 L

Algebra ->  Inequalities -> SOLUTION: Choose a possible dimensions of a rectangle with a width= 2x-1 and length= 3x+2 so that its area is greater than 153cm² a. W= 13 L= 15 b. W= 11 L= 20 c. W= 11 L= 21 d. W= 13 L      Log On


   

Question 1170391: Choose a possible dimensions of a rectangle with a width= 2x-1 and length= 3x+2 so that its area is greater than 153cm²
a. W= 13 L= 15
b. W= 11 L= 20
c. W= 11 L= 21
d. W= 13 L= 22
#My answer would be 10cm by 18cm, is it correct? or not?
Found 2 solutions by MathLover1, Clanther:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

a rectangle with a width=+2x-1 and length=+3x%2B2+so that its area is greater than 153cm%5E2
area
A=%283x%2B2%29%28+2x-1%29+
A=6+x%5E2+%2B+x+-+2
given area is greater than 153cm%5E2
6+x%5E2+%2B+x+-+2%3E153
6+x%5E2+%2B+x+-+2-153%3E0
6+x%5E2+%2B+x+-155%3E0
6+x%5E2+%2B31x-30x+-155%3E0
%286+x%5E2+-30x%29%2B%2831x+-155%29%3E0
6x%28+x+-5%29%2B31%28x+-5%29%3E0
%28x+-+5%29+%286+x+%2B+31%29%3E0
solutions:
x%3E5
or
x+%3E-31%2F6..disregard negative solution
find the width and length if x%3E5
let x=6, first number greater than 5
a width=+2x-1=2%2A6-1=11
and length=+3x%2B2=3%2A6%2B2=20


so, answer is: b. W=+11 L=+20


Answer by Clanther(6) About Me  (Show Source):
You can put this solution on YOUR website!
letter b w=11 l=20
Equation:
(3x+2)(2x-1)> 153
One of the roots that is positive is 5
x>5 , the first value that is greater than 5 is 6
width:
2x-1
2(6)-1= 11cm
length:
3x+2
3(6)+2= 20cm