Question

Each chef at "Sushi Emperor" prepares 15 regular rolls and 20 vegetarian rolls daily. On Tuesday, each customer ate 2 regular rolls and 3 vegetarian rolls. By the end of the day, 4 regular rolls and 1 vegetarian roll remained uneaten. How many chefs and customers were in "Sushi Emperor" on Tuesday?

Answer 1

9514 1404 393

Answer:

• 2 chefs
• 13 customers

Step-by-step explanation:

Let x and y represent the numbers of chefs and customers, respectively. We can write equations for the different kinds of rolls made and eaten:

  15x -2y = 4 . . . . . 4 regular rolls were uneaten

  20x -3y = 1 . . . . . 1 vegetarian roll was uneaten

Using the "cross multiplication method", the solution can be found to be ...

  Δ1 = (15)(-3) -(20)(-2) = -5

  Δ2 = (-2)(-1) -(-3)(-4) = -10

  Δ3 = (-4)(20) -(-1)(15) = -65

1/Δ1 = x/Δ2 = y/Δ3, so we have ...

  x = Δ2/Δ1 = -10/-5 = 2 . . . . there were 2 chefs

  y = Δ3/Δ1 = -65/-5 = 13 . . . there were 13 customers

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Additional comment

The version of the "cross multiplication method" I use for solving two linear equations is this. Write the equations in general form. List the coefficients in two rows, repeating the first column so there are 4 columns. Form "cross products" of the coefficients taking columns pairwise. Use these cross products to find the variable values.

In general form, these equations are ...

• 15x -2y -4 = 0
• 20x -3y -1 = 0

Then the rows of coefficients are ...

  [ 15, -2, -4, 15 ]

  [ 20, -3, -1, 20 ]

The "cross multiplication" subtracts the product on the up-diagonal from the product on the down-diagonal. If we label the results Δ1, Δ2, Δ3, we get the results shown above. These "cross products" are used in the equation ...

  1/Δ1 = x/Δ2 = y/Δ3

to find the values of x and y.

Some videos I have seen of this method write the two rows of coefficients shifted to the left, and use a different final equation for the solution. The result is the same. I find the extra shifting to make the process confusing and difficult to remember.

I prefer this method when there is no simple way to do substitution or elimination. In the end, it takes slightly fewer math operations than would be required by either of those methods in such cases.

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A graphing calculator provides another simple way to find the solution.