Given:
• Volume of small containers = 2.5 m³
,
• Volume of large containers = 4 m³
,
• Number of containers in the shipment = 140
,
• Total volume = 440 m³
Let's find the number of the size of each container used.
Let s represent the number of small containers
Let L represent the number of large containers.
• Equation for number of containers:
s + L = 140
• Equation for total volume:
2.5s + 4L = 140
Hence, we have the set of equations:
s + L = 140...........................equation 1
2.5s + 4L = 440...................equation 2
Let's solve the set of equations simultaneously using substitution methos.
Rewrite equation 1 for L:
L = 140 - s...........................equation 3
Substitute (140 -s) for L in equation 2:
2.5s + 4(140 - s) = 440
Apply distributive property:
2.5s + 4(140) + 4(-s) = 440
2.5s + 560 - 4s = 440
Combine like terms:
2.5s - 4s + 560 = 440
-1.5s + 560 = 440
Subtract 560 from both sides:
-1.5s + 560 - 560 = 440 - 560
-1.5s = -120
Divide both sides by -1.5:
Substitute 80 for s in either of the equations.
Take equation 3.
L = 140 - s
L = 140 - 80
L = 60.
Therefore, we have the solutions:
s = 80 and L = 60
Therefore, the company used 80 small containers and 60 large containers
