Answer:
Mileage = $71
Loading/Unloading Fee: -$980
Explanation:
The question has incomplete details. The complete question can be found online. The given parameters are:
Situation A
[tex]Helpers = 6[/tex]
[tex]Distance =95[/tex]
[tex]Cost = 865[/tex]
Situation B
[tex]Helpers = 5[/tex]
[tex]Distance = 80[/tex]
[tex]Cost = 780[/tex]
Solving (a): The equation.
Let
x = loading/unloading fee
y = mileage fee
For situation A, the equation will be:
[tex]Cost = 6 * x + 95 * y[/tex]
[tex]Cost = 6 x + 95y[/tex]
[tex]6 x + 95y = 865[/tex]
For situation B, the equation will be:
[tex]Cost = 5* x + 80*y[/tex]
[tex]Cost = 5x + 80y[/tex]
[tex]5x + 80y = 780[/tex]
So, the equations are:
[tex]6 x + 95y = 865[/tex]
[tex]5x + 80y = 780[/tex]
Solving (b): The solution
[tex]6 x + 95y = 865[/tex] --- (1)
[tex]5x + 80y = 780[/tex] --- (2)
Multiply (1) by 5 and multiply (2) by 6 to eliminate x
[tex]5 * [[/tex] [tex]6 x + 95y = 865[/tex][tex]][/tex]
[tex]30x + 475y = 4325[/tex]
[tex]6*[[/tex][tex]5x + 80y = 780[/tex][tex]][/tex]
[tex]30x + 480y = 4680[/tex]
Subtract the resulting equations
[tex]30x - 30x + 475y - 480y = 4325 - 4680[/tex]
[tex]-5y = -355[/tex]
Divide both sides by -5
[tex]y = 71[/tex]
Substitute [tex]y = 71[/tex] in (1)
[tex]6 x + 95y = 865[/tex]
[tex]6x + 95 * 71 = 865[/tex]
[tex]6x + 6745 = 865[/tex]
Collect like terms
[tex]6x =- 6745 + 865[/tex]
[tex]6x =-5880[/tex]
Divide both sides by 6
[tex]x =-980[/tex]
The solution is correct but does not represent real life situation because x should be positive
