The sum of the length, width, and height of a rectangular box is 75 cm. The length is twice the sum of the width and height, and twice the width exceeds the height by 5cm. Find the dimensions.
If L = length, W = width, H = height
L = 2(W+H)
2W = H+5
So L = 2W + 2H = H+5 + 2H = 3H+5
W = (H+5)/2
Now: L + W + H = 75
(3H+5) + (H+5)/2 + H = 75
2(3H+5) + (H+5) + 2H = 150
6H + 10 + H + 5 + 2H = 150
9H = 135
H = 15
L = 3H + 5 = 45+5 = 50
W = (H+5)/2 = 20/2 = 10
1) L + w + h = 75
2) L = 2(w + h)
3) 2w = h + 5 ---> h = 2w - 5
Substituting from 3) for h in 2), we get
L = 2(w + 2w -5)
L = 2(3w -5)
L = 6w -10
Substitute this for L in equation 1).
6 w - 10 + w + h = 75 ---> 7w +h = 85 ---> h = 85 - 7w
Substitute for h in equation 3)
85 - 7w = 2w - 5 ---> 90 = 9w ---> w = 10
Equation 3) has only 2 unknowns, with w now known, so solve for h:
2(10) = h + 5
h = 15
Now substitute 15 for h and 10 for w in 1)
L + 15 + 10 = 75
L = 50
Now check with the words using length is 50 cm, width is 10cm, and height is 15 cm:
Is the sum of 50 cm,10 cm, and 15 cm = to 75 cm?
Is 50 cm twice the sum of 15 cm and 10 cm?
Does twice 10 cm exceed 15cm by 5 cm?
First relate the dimensions
l+w+h=75
l = 2(w+h)
2w = h+5
we can say that h = 2w - 5 and use it in the second relation.
l = 2(w+2w-5) = 6w-10
since we already got a relation for l to w and h to w, we uses the first relation.
6w -10 + w + 2w - 5 = 75
9w=90
w=10
then use this value of w for the other equations.
h=15
l=50
Comments
If L = length, W = width, H = height
L = 2(W+H)
2W = H+5
So L = 2W + 2H = H+5 + 2H = 3H+5
W = (H+5)/2
Now: L + W + H = 75
(3H+5) + (H+5)/2 + H = 75
2(3H+5) + (H+5) + 2H = 150
6H + 10 + H + 5 + 2H = 150
9H = 135
H = 15
L = 3H + 5 = 45+5 = 50
W = (H+5)/2 = 20/2 = 10
1) L + w + h = 75
2) L = 2(w + h)
3) 2w = h + 5 ---> h = 2w - 5
Substituting from 3) for h in 2), we get
L = 2(w + 2w -5)
L = 2(3w -5)
L = 6w -10
Substitute this for L in equation 1).
6 w - 10 + w + h = 75 ---> 7w +h = 85 ---> h = 85 - 7w
Substitute for h in equation 3)
85 - 7w = 2w - 5 ---> 90 = 9w ---> w = 10
Equation 3) has only 2 unknowns, with w now known, so solve for h:
2(10) = h + 5
h = 15
Now substitute 15 for h and 10 for w in 1)
L + 15 + 10 = 75
L = 50
Now check with the words using length is 50 cm, width is 10cm, and height is 15 cm:
Is the sum of 50 cm,10 cm, and 15 cm = to 75 cm?
Is 50 cm twice the sum of 15 cm and 10 cm?
Does twice 10 cm exceed 15cm by 5 cm?
First relate the dimensions
l+w+h=75
l = 2(w+h)
2w = h+5
we can say that h = 2w - 5 and use it in the second relation.
l = 2(w+2w-5) = 6w-10
since we already got a relation for l to w and h to w, we uses the first relation.
l+w+h=75
6w -10 + w + 2w - 5 = 75
9w=90
w=10
then use this value of w for the other equations.
h=15
l=50