The width of a rectangular box is twice the height. Twice the length of the box exceeds the sum of the width and the height by 5in. Given that the sum of the length, width, and height is 16in, find the length, width, and height.
I got,
w=2h
2L=w+h+5
L+w+h=16
I don't know if the problems are right. I would like to know if they are or not. I can do the rest of the problem, I'm just unsure if I got the right problems out of the story problem.
They look correct to me!
ok, I'll take your word for it
Your initial setup of the problem seems correct. Let's go through it step by step to confirm:
Let:
Width of the box = w
Height of the box = h
Length of the box = L
From the problem statement, we have the following information:
1) "The width of a rectangular box is twice the height": This can be represented as w = 2h.
2) "Twice the length of the box exceeds the sum of the width and the height by 5in": Mathematically, this can be expressed as 2L = w + h + 5.
3) "The sum of the length, width, and height is 16in": This can be written as L + w + h = 16.
Based on these three equations, you correctly set up the equations as follows:
w = 2h (Equation 1)
2L = w + h + 5 (Equation 2)
L + w + h = 16 (Equation 3)
So, your initial setup and equations are correct. Now you can proceed with solving these equations to find the values of L, w, and h.