Question

3. Jaden received these two estimates from a moving company. Write and solve a

system of equations to determine the hourly loading/unloading fee and the mileage

charge for the truck rental.

Situation A: He hires 6 helpers to load and unload the truck and

travels 95 miles on back roads for a total cost of $865.

Answer 1

Mileage = $71

Loading/Unloading Fee: -$980

Step-by-step explanation:

The question has incomplete details. The complete question can be found online. The given parameters are:

Situation A

`Helpers = 6`

`Distance =95`

`Cost = 865`

Situation B

`Helpers = 5`

`Distance = 80`

`Cost = 780`

Solving (a): The equation.

Let

x = loading/unloading fee

y = mileage fee

For situation A, the equation will be:

`Cost = 6 * x + 95 * y`

`Cost = 6 x + 95y`

`6 x + 95y = 865`

For situation B, the equation will be:

`Cost = 5* x + 80*y`

`Cost = 5x + 80y`

`5x + 80y = 780`

So, the equations are:

`6 x + 95y = 865`

`5x + 80y = 780`

Solving (b): The solution

`6 x + 95y = 865` --- (1)

`5x + 80y = 780` --- (2)

Multiply (1) by 5 and multiply (2) by 6 to eliminate x

`5 * [`
`6 x + 95y = 865`
`]`

`30x + 475y = 4325`

`6*[`
`5x + 80y = 780`
`]`

`30x + 480y = 4680`

Subtract the resulting equations

`30x - 30x + 475y - 480y = 4325 - 4680`

`-5y = -355`

Divide both sides by -5

`y = 71`

Substitute
`y = 71` in (1)

`6 x + 95y = 865`

`6x + 95 * 71 = 865`

`6x + 6745 = 865`

Collect like terms

`6x =- 6745 + 865`

`6x =-5880`

Divide both sides by 6

`x =-980`

The solution is correct but does not represent real life situation because x should be positive