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Difficulty: HardLinear Equations in One and Two Variables

A technology manufacturing company produces two server models: Model Alpha and Model Beta. The manufacturing cost of one Model Alpha server is 400morethantwicethemanufacturingcostofoneModelBetaserver.Lastmonth,thecompanyproduced15ModelAlphaserversand40ModelBetaservers,incurringatotalmanufacturingcostof400 more than twice the manufacturing cost of one Model Beta server. Last month, the company produced 15 Model Alpha servers and 40 Model Beta servers, incurring a total manufacturing cost of 76,000. What is the manufacturing cost, in dollars, of one Model Alpha server?

Answer: 2400 dollars

Answer

The manufacturing cost of one Model Alpha server is $2,400.
By defining the costs as A=2B+400A = 2B + 400 and setting up the total cost equation 15A+40B=76,00015A + 40B = 76,000, substituting AA yields 70B=70,00070B = 70,000, giving B=1,000B = 1,000. Substituting B=1,000B = 1,000 back into A=2B+400A = 2B + 400 gives A=2,400A = 2,400.

Step-by-Step Solution

1
Define variables and construct the linear relation equation.
A=2B+400A = 2B + 400
The cost of one Model Alpha server (AA) is 400morethantwicethecostofModelBeta(400 more than twice the cost of Model Beta ( B$).
2
Formulate the total production cost equation.
15A+40B=76,00015A + 40B = 76,000
15 units of Model Alpha and 40 units of Model Beta sum to a total cost of $76,000.
3
Substitute AA into the total cost equation and solve for BB.
15(2B+400)+40B=76,000  ⟹  70B+6,000=76,000  ⟹  B=1,00015(2B + 400) + 40B = 76,000 \implies 70B + 6,000 = 76,000 \implies B = 1,000
Substituting reduces the system to a single linear equation in one variable.
4
Calculate the value of AA.
A=2(1,000)+400=2,400A = 2(1,000) + 400 = 2,400
Substitute B=1,000B = 1,000 back into the equation for AA.

Key Concept

Solving Systems of Linear Equations by Substitution
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Topics

Linear Equations in One and Two Variables

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