SOLUTION: We're trying to help our soon with his math homework but we're totally stuck one this problem: A store has a sale on almonds, pecans and pistachios. One lb of almonds, one lb

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Question 615967: We're trying to help our soon with his math homework but we're totally stuck one this problem:
A store has a sale on almonds, pecans and pistachios. One lb of almonds, one lb of pecans and one lb of pistachios cost $12. Two lbs of almonds and three lbs of pecans cost $16. Three lbs of pecans and two lbs pistachios cost $24. Find the price of each kind of nut.

Thank you in advance for any help.
Found 2 solutions by scott8148, Theo:
Answer by scott8148(6628) (Show Source):
You can put this solution on YOUR website!
x=almond, y=pecan, z=pistachio

x + y + z = 12 ___ A

2x + 3y = 16 ___ B

3y + 2z = 24 ___ C

eliminating x ___ 2A - B ___ -y + 2z = 8 ___ D

eliminating y ___ 3D + C ___ 8z = 48

solve for z, then substitute back to find y and x

Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
your answer is:
a = 2
b = 4
c = 6

a = pounds of almonds
b = pounds of pecans
c = pounds of pistachios

you have 3 equations that need to be solved simultaneously.

they are:
a + b + c = 12 (first equation)
2a + 3b = 16 (second equation)
3b + 2c = 24 (third equation)

take the first and second equation and solve them together to eliminate b.
multiply the first equation by 3 to get:
3a + 3b + 3c = 36
subtract second equation of 2a + 3b = 16 from it.
you get:
3a - 2a = a
3b - 3b = 0
3c - 0 = 3c
36 - 16 = 20
you are left with:
a + 3c = 20 (fourth equation)

take the first and third equations and solve them together to eliminate b.
multiply the first equation by 3 to get:
3a + 3b + 3c = 36
subtract third equation of 3b + 2c = 24 from it.
you get:
3a - 0 = 3a
3b - 3b = 0
3c - 2c = c
36 - 24 = 12
you are left with:
3a + c = 12 (fifth equation)

solve the fourth and fifth equations simultaneously to eliminate a.
you start with:
a + 3c = 20 (fourth equation)
3a + c = 12 (fifth equation)

multiply the fourth equation by 3 to get:
3a + 9c = 60
subtract the fifth equation of 3a + c = 12 from it.
you get:
3a - 3a = 0
9c - c = 8c
60 - 12 = 48
you are left with:
8c = 48
divide both sides of this equation by 8 to get:
c = 6

take the third equation to get:
3b + 2c = 24
substitute 6 for c in that equation and solve for b.
you get:
3b + 2*6 = 24 which becomes:
3b + 12 = 24 which becomes:
3b = 12 which becomes:
b = 4

take the first equation to get:
a + b + c = 12
substitute 6 for c and 4 for b to get:
a + 4 + 6 = 12
solve for a to get:
a = 2

your solutions are:
a = 2
b = 4
c = 6

substitute in all 3 original equations to confirm these solutions are good.

your original 3 equations are:
a + b + c = 12 (first equation)
2a + 3b = 16 (second equation)
3b + 2c = 24 (third equation)
after substitution of 2 for a and 4 for b and 6 for c, these equations become:
2 + 4 + 6 = 12
4 + 12 = 16
12 + 12 = 24
all these equation are true after substituting the values of a,b,c in them so the solutions are good because they solve all 3 equations simultaneously.