As a restaurant owner there are many decisions that you need to make on a daily basis, such as where to keep inventory levels. You wish to replenish your stock of dishes by purchasing 250 sets for your restaurant. You have two dish design from which to choose. One design costs $20 per set and the other $45 per set. If you only have $6,800 to spend, how many of each design should you order?

The algebraic way:

Let
x = number of $45 sets, and
250-x = number of $20 sets

45x+20(250-x)=6800
25x = 6800-5000
x=1800/25=72
Therefore you would buy 72 sets at $45 and (250-72)=178 sets at $20.

Check the answers.

The layman's way (no calculator, no pencil required):

Buy all 250 sets at $20, that leaves $6800-5000=1800 to spare. For every set we change from $20 to $45, we pay an extra $45-20=$25. So we can change $1800/$25=72 sets.
The answer is therefore 72 sets at $45 and 250-72=178 sets at $20.

7x4

To determine how many of each design you should order, you can set up a system of equations.

Let's represent the number of sets of the cheaper design as "x" and the number of sets of the more expensive design as "y".

The cost constraint can be expressed as:

20x + 45y = 6,800

The total number of sets constraint can be expressed as:

x + y = 250

Now we can solve this system of equations to find the values of x and y.

First, we can rearrange the second equation to solve for x:

x = 250 - y

Substitute this expression for x in the first equation:

20(250 - y) + 45y = 6,800

Now simplify and solve for y:

5,000 - 20y + 45y = 6,800
25y = 1,800
y = 72

Now substitute the value of y back into the second equation to find x:

x + 72 = 250
x = 250 - 72
x = 178

Therefore, you should order 178 sets of the cheaper design and 72 sets of the more expensive design.