Question

Use the given cost table for the same product from two different companies to create alinear system. Then solve the system to determine when the cost of the product will be thesame and what the price will be.Let k(n) represent the cost for Company 1 and let m(n) represent the cost of Company 2,where n is the number of tons of road salt.Road Salt (tons)Company 1Company 25$1,525$2,35010$2,850$3,40015$4,175$4,450n +k(n) =m(n) =n+Both Company 1 and Company 2 charge $fortons of road salt.

Answer 1

In this case, we'll have to carry out several steps to find the solution.

Step 01:

Data

k(n) = cost of Company 1

m(n) = cost of Company 2

n = number of tons of road salt.

linear system = ?

Step 02:

linear system

Company 1

k(n) =

slope equation

`m\text{ = }(y2-y1)/(x2-x1)=(2850-1525)/(10-5)=(1325)/(5)=265`

Point-slope form of the line

(y - y1) = m (x - x1)

(y - 1525) = 265 (x - 5)

y = 265x - 1325 + 1525

y = 265x + 200

k(n) = 265n + 200

Company 2

m(n) =

slope equation

`m\text{ = }(y2-y1)/(x2-x1)=(3400-2350)/(10-5)=(1050)/(5)=210`

(y - y1) = m (x - x1)

(y - 2350) = 210 (x - 5)

y = 210x - 1050 + 2350

y = 210x + 1300

m(n) = 210n + 1300

Step 03:

cost of the product will be the same

k(n) = m (n)

265n + 200 = 210n + 1300

265n - 210n = 1300 - 200

55n = 1100

n = 1100 / 55 = 20

k(n) = 265n + 200

k(20) = 265*20 + 200 = 5500

The full solution is:

The linear system is:

k(n) = 265n + 200

m(n) = 210n + 1300

n = 20 tons of road salt

The cost will be $5500 for 20 tons of road salt (Company 1 and Company 2) (Do you have questions about the steps to solve the exercise or about the final solution?