How do you solve this problem?
Topic: Steps to solving a word problem
May 25, 2019 / By Candida Question:
Please help, I'm doing algebra 2 summer work and it's stressing me out. I need help solving this word problem: A room's length is 3 feet less than twice its width. The area of the room is 135 square feet. What are the room's dimensions? I'm really confused because I tried substituting the variable of 'l' with 2w-3 so I could set up an equation that read 2w-3*w=135 but when I solved it I came out with radical 69 equal to w and when I went back and plugged it into the equation it didn't work. It would be helpful if you showed me the steps so i knew how to do it, thank you.
Best Answers: How do you solve this problem?
Amber | 9 days ago
You must have two independent equations to solve for the two variables, L (length) and W (width):
A = LW = 135
L = 2W - 3
Substitute L into the first equation and solve for W:
(2W - 3)W = 135
2W² - 3W - 135 = 0
(2W + 15)(W - 9) = 0
Therefore, W = -15/2 and W = 9. Since the width cannot be negative, then W = 9 is the only solution. Substitute W = 9 into the second equation and solve for L:
L = 2W - 3 = 2(9) - 3 = 15
Thus, the room is 15 feet long and 9 feet wide.
👍 106 | 👎 9
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Originally Answered: Can anyone help me in solving this math problem? I have spent 4 hours to solve it but couldnt solve?
I did trial and error and came up with 900.
900 is divisible by 12, 15 and 18. The answers are 75, 60 and 50, respectively.
The square root of 900 is 30.
You know the number has to be a whole number squared, since they can be arranged in a square, and it also has to be divisible by 12, 15 and 18.
I started at 18^2 and went up, first dividing by 18, and if the number squared divided by that was a whole number I checked it with the other two numbers... ended up getting to 30.
So the equations I used were x^2/18, x^2/15 and x^2/12 and all had to give me whole numbers. x has to be a positive integer and the number of students was x^2.
Hope that makes sense.
let the length be L and width be W
L is 3 ft less than twice W
L = 2w - 3...........(1)
Area = L x W = 135
(2w - 3) x W = 135
2W² - 3W =135
2W² - 3W - 135 = 0
2W² + 15W - 18W - 135 = 0
W(2W + 15) - 9(2W + 15) = 0
(2W + 15) (W - 9) = 0
Therefore 2W + 15 = 0 OR W - 9 = 0
neglecting the negative value W = 9 ft
L = 2 x 9 - 3 = 15
Dimensions are Length = 15 ft and width = 9 ft
Area = L x W = 15 x 9 = 135 which tallies with the data
👍 30 | 👎 5
Let's call the width w.
"A room's length is 3 feet less than twice its width." So the length is 2w - 3
"The area of the room is 135 square feet." So w * l = 135 sq ft
w * (2w - 3) = 135
2w^2 - 3w = 135
2w^2 - 3w - 135 = 0
w = (-(-3) +/- sqrt((-3)^2 - 4 * 2 * -135)) / (2 * 2)
w = (3 +/- sqrt(9 + 1080)) / 4
w = (3 +/- sqrt(1089)) / 4
w = (3 +/- 33) / 4
w = 36/4 or -30/4
w = 9 or -7.5
Obviously w cannot equal 9.5, so w = 9 and l = 2 * 9 - 3 = 15.
👍 22 | 👎 1
Taking your equation
using Quadratic equation:
w=(3±√9+1080) / (4)
w=(3±33) / (4)
Width cant be negative so, we will take only:
w=(3+33) / (4)
Hence, the width of the room will be 9 feet and the length will be 15 feet.
👍 14 | 👎 -3
let the width = w
now, length* width = area
i.e. (2w-3)* w= 135
i.e. 2w^2 -3w = 135
i.e. 2w^2 - 3w - 135=0
i.e. 2w^2-18w+15w -135=0
either 2w+15=0 i.e. w= -15/2 which is impossible because width never -ve.
or, w-9=0 i.e. w=9
therefore, width= 9 ft and length= (2*9 - 3)ft.= 15 ft.
👍 6 | 👎 -7
You wrote this in your question yet, it is not just a typo on your part, right?
Solve that and you get the width.
It looks like you will need the quadratic formula, which is so long winded that I am not going to do it here. Make sure you take the positive value of w,
and remember that
👍 -2 | 👎 -11
Originally Answered: I don't seem to solve this proble at all, can you please help me solve calculus problem?
You need to draw it first so you can visualize. I can't do that on a computer. However, it is basically one parabola over another, with the origin of the 2nd one at (0,2) and of course y = x² has its origin at (0,0). Then you spin that around the y axis to see the solid that is formed. It will look like two bullets with the bases of the bullets touching.
dV = πr² dr the infitesmal volume is the area of a circle times its "thickness" dr.
However, r is going to vary from zero at the very bottom to 1 at the x axis (0,0) to 0 again at (0,2) and the only way we can describe the way r varies is to express it as a function of x namely y = x²
However this is pretty easy by looking at the symmetry of the solid. You can cut it in half about a plane parallel to y = 1 so that means figure out what ½ of the volume is and multiply by 2 to get the total volume.
dV = πr² dr
were r = x² dr = 2x dx
dV = 2&pix³ dx
V = ∫ dV = ∫ 2&pix³ dx as x goes from 0 to 1
V = 2 (x^4)/4 evaulated between 0 and 1
V = ½ x^4 evaluated between 0 and 1
V = ½(0)^4 - ½(1)^4 = -½
so that is the "negative" volume below the plane parallel to y =1
so double it to get a final answer of 1