How much does it cost to run Unit A per hour? 0.19 0.22 0.24 0.27

Question

ScanSolve AI Tutor · Accepted Answer

To find the cost per hour to run Unit A, we need to set up and solve a system of linear equations based on the provided data.

Let $C_A$ , $C_B$ , and $C_C$ be the cost per hour for Unit A, Unit B, and Unit C, respectively.

Formulate the equations from the given data.
Based on the "Number of Hours Worked" and "Total Air Conditioning Electric Costs" tables, we can write an equation for each month:
For June:
$184 C_A + 207 C_B + 276 C_C = 142.60$ (Equation 1)
For July:
$174 C_A + 198 C_B + 264 C_C = 135.96$ (Equation 2)
For August:

$168 C_A + 189 C_B + 242 C_C = 128.30$ (Equation 3)
Create new equations by subtracting the original equations.
Subtract Equation 2 from Equation 1:
$(184 - 174)C_A + (207 - 198)C_B + (276 - 264)C_C = 142.60 - 135.96$
$10 C_A + 9 C_B + 12 C_C = 6.64$ (Equation 4)

Subtract Equation 3 from Equation 2:
$(174 - 168)C_A + (198 - 189)C_B + (264 - 242)C_C = 135.96 - 128.30$
$6 C_A + 9 C_B + 22 C_C = 7.66$ (Equation 5)
Eliminate $C_B$ from the new equations.
Subtract Equation 4 from Equation 5:
$(6 - 10)C_A + (9 - 9)C_B + (22 - 12)C_C = 7.66 - 6.64$
$-4 C_A + 10 C_C = 1.02$ (Equation 6)
Express $C_C$ in terms of $C_A$ .
From Equation 6:
$10 C_C = 1.02 + 4 C_A$
$C_C = 0.102 + 0.4 C_A$
Substitute $C_C$ into Equation 4 to get an equation with $C_A$ and $C_B$ .
$10 C_A + 9 C_B + 12 (0.102 + 0.4 C_A) = 6.64$
$10 C_A + 9 C_B + 1.224 + 4.8 C_A = 6.64$
$14.8 C_A + 9 C_B = 6.64 - 1.224$
$14.8 C_A + 9 C_B = 5.416$ (Equation 7)
Substitute $C_C$ into Equation 1 to get another equation with $C_A$ and $C_B$ .
$184 C_A + 207 C_B + 276 (0.102 + 0.4 C_A) = 142.60$
$184 C_A + 207 C_B + 28.152 + 110.4 C_A = 142.60$
$294.4 C_A + 207 C_B = 142.60 - 28.152$
$294.4 C_A + 207 C_B = 114.448$ (Equation 8)
Solve the system of equations for $C_A$ and $C_B$ .
We have:
$14.8 C_A + 9 C_B = 5.416$ (Equation 7)
$294.4 C_A + 207 C_B = 114.448$ (Equation 8)

To eliminate $C_B$ , multiply Equation 7 by $\frac{207}{9} = 23$ :
$23 \times (14.8 C_A + 9 C_B) = 23 \times 5.416$
$340.4 C_A + 207 C_B = 124.568$ (Equation 9)

Subtract Equation 8 from Equation 9:
$(340.4 - 294.4) C_A + (207 - 207) C_B = 124.568 - 114.448$
$46 C_A = 10.12$
$C_A = \frac{10.12}{46}$
$C_A = 0.22$

The cost to run Unit A per hour is $0.22.

The final answer is \boxed{\0.22}$.

How much does it cost to run Unit A per hour? 0.19 0.22 0.24 0.27

Answer

Step-by-step solution

Formulate the equations from the given data.

Create new equations by subtracting the original equations.

Eliminate $C_B$ from the new equations.

Express $C_C$ in terms of $C_A$ .

Substitute $C_C$ into Equation 4 to get an equation with $C_A$ and $C_B$ .

Substitute $C_C$ into Equation 1 to get another equation with $C_A$ and $C_B$ .

Solve the system of equations for $C_A$ and $C_B$ .

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How much does it cost to run Unit A per hour? 0.19 0.22 0.24 0.27

Answer

Step-by-step solution

Formulate the equations from the given data.

Create new equations by subtracting the original equations.

Eliminate CBC_BCB​ from the new equations.

Express CCC_CCC​ in terms of CAC_ACA​.

Substitute CCC_CCC​ into Equation 4 to get an equation with CAC_ACA​ and CBC_BCB​.

Substitute CCC_CCC​ into Equation 1 to get another equation with CAC_ACA​ and CBC_BCB​.

Solve the system of equations for CAC_ACA​ and CBC_BCB​.

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Still stuck on this one?

More Mathematics Questions

Eliminate $C_B$ from the new equations.

Express $C_C$ in terms of $C_A$ .

Substitute $C_C$ into Equation 4 to get an equation with $C_A$ and $C_B$ .

Substitute $C_C$ into Equation 1 to get another equation with $C_A$ and $C_B$ .

Solve the system of equations for $C_A$ and $C_B$ .