### Supper with Skiing

A ski resort offers two package plans. Plan A gives one person four nights lodging and three dinners for\$326. Plan B gives one person five nights lodging and four dinners for \$413. Assuming that the costs per night for lodging anddinner are the same for both plans, how much does one dinner cost?

(Source: Mathematics Teaching in the Middle School, Mar-Apr 1996).

### The Solution

We can set this problem up as a system of equations, and then solve for the uknown variables. We are given two distinct situations, the first is a skiing package offering four nights of lodging and three dinners for \$326. If we allow l= cost of lodging per night and d= cost of each dinner, then we can write the following equation:

4l + 3d = 326

Using the same idea, we can write an equation for plan B, which porvides five nights of lodging and four dinners for \$413:

5l + 4d = 413

Next we can solve for one of the variables in one of the equations, and use that solution to plug into the other equation and then solve.

5l +4d = 413

5l = 413-4d

l = (413-4d)/5

Now that we have solved for l in terms of d, we can plug the solution for l into the first equation to solve for d:

## 4*[(413-4d)/5] + 3d = 326

330.4 - 3.2d + 3d = 326

330.4 - 0.2d = 326

4.4 = 0.2d

22 = d

Now we have solved for the price of each dinner, and we can use that solution to find the price of each nights lodging by simply plugging the price of dinner into the original equations that we have. Solving for the price of lodging in both equations should produce the same answer:

Plan A
4l + 3(22) = 326

4l = 260

l = 65

Plan B

5l + 4(22) = 413

5l = 325

l = 65

Thus we get a final solution telling us that each night of lodging costs \$65.00 and each dinner costs \$22.00

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