Four watermelons and three cantaloupes weigh 44 pounds. All watermelons weigh the same and all cantaloupes weigh the same. What is the weight of two watermelons and one cantaloupe?
2006-07-14 15:06:31 · 5 answers · asked by Jack H 2 in Science & Mathematics ➔ Mathematics
This is an algebra problem.
Watermelon weight will be W
Cantaloupe weight will be C
Equation 1: 3W + 2C = 32 pounds
Equation 2: 4W + 3C = 44 pounds
2 equations, 2 unknowns, solve for W and C:
2C = 32 pounds - 3W
C = 16 pounds - 3/2 W
4W + 3(16 pounds - 3/2W) = 44 pounds
4W + 48 pounds - 9/2W = 44 pounds
4 pounds = 1/2W
W = 8 pounds
C = 16 pounds - 3/2(8 pounds) = 4 pounds
Question is weight of 2W + 1C = 2(8 pounds) + 4 pounds = 20 pounds
The answer is 20 pounds.
2006-07-14 15:20:14 · answer #1 · answered by SkyWayGuy 3 · 0⤊ 0⤋
Adding one watermelon and one cantaloupe added 12 pounds so one of each = 12 pounds. Subtract twelve from 32 since you took away one of each.
Leaves you 20 pounds
2006-07-14 22:29:00 · answer #2 · answered by Anonymous · 0⤊ 0⤋
3x + 2y = 32
4x + 3y = 44
(3x + 2y) + (4x + 3y) = 32 + 44
3x + 2y + 4x + 3y = 76
7x + 5y = 76
5y = -7x + 76
y = (-7/5)x + (76/5)
3x + 2y = 32
3x + 2((-7/5)x + (76/5)) = 32
3x + (-14/5)x + (152/5) = 32
15x - 14x + 152 = 160
x + 152 = 160
x = 8
y = (-7/5)x + (76/5)
y = (-7/5)(8) + (76/5)
y = (-56/8) + (76/5)
y = (-56 + 76)/5
y = (20/5)
y = 4
Watermelons = $8
Cantaloupes = $4
2006-07-15 13:33:50 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
3W + 2C = 32
4W + 3C = 44
mult top eq by (3) ---> 9W + 6C = 96
mult btm eq by (2) ----> 8W + 6C = 88
subtract
1W = 8
plug in 8 for W in orig eq 3(8) + 2C = 32
2C = 8
C= 4
2006-07-14 22:55:39 · answer #4 · answered by eddiervc 3 · 0⤊ 0⤋
20 pounds.
2006-07-14 22:09:41 · answer #5 · answered by shay p 2 · 0⤊ 0⤋