SOLUTION: Linear systems word problem:
"The cost of six pennants and nine balls is $48. The cost of three balls is $1 more than the cost of one pennant. What are the costs of one pennant an
Question 701258: Linear systems word problem:
"The cost of six pennants and nine balls is $48. The cost of three balls is $1 more than the cost of one pennant. What are the costs of one pennant and one ball?"
I need help writing two equations and can it be solved using elimination? If not,can you solve it for me using substitution? Thanks! Found 2 solutions by MathLover1, stanbon:Answer by MathLover1(20850) (Show Source):
You can put this solution on YOUR website!
let the cost of pennants be and the cost of balls :
if the cost of six pennants and nine balls is $ we can write it as
......eq. 1
if the cost of three balls is $ more than the cost of one pennant, than we write it as
....eq. 2
now we have system to solve:
......eq. 1 ....eq. 2
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start with ....eq. 2 and solve for .....substitute it in eq. 1
......solve for ....the cost of pennants
now find ......the cost of balls
You can put this solution on YOUR website! The cost of six pennants and nine balls is $48. The cost of three balls is $1 more than the cost of one pennant. What are the costs of one pennant and one ball?"
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Value Equations:
6p + 9b = 48
3b = p + 1
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Substitution:
p = 3b-1
Substitute for "p" and solve for "b":
6(3b-1) + 9b = 48
18b - 6 + 9b = 48
27b = 54
b = $2.00 (cost of 1 ball)
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Solve for "p":
p = 3b-1
p = 3*2-1
p = $5.00 (cost of one pennant)
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Elimination:
6p + 9b = 48
3b = p + 1
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Rearrange
3b + 2p = 16
3b - p = 1
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Subtract and solve for "p"
3p = 15
p = $5.00 (pennant)
Solve for "b":
3b - p = 1
3b - 5 = 1
3b = 6
b = $2.00 (ball)
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Cheers,
Stan H.
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