A restaurant owner plans to use x tables seating 4, y tables seating 6, and z tables seating 8, for a total of 20 tables. When fully occupied, the tables seat 108 customers. If only half of the x tables, half of the y tables, and one-fourth of the z tables are used, each fully occupied, then 46 customers will be seated. Find x, y, and z

x = 10, y = 6 and z = 4

Step-by-step explanation:

From the question, we can form simultaneous equations of 3 variables.

Since the total tables is 20, then

x + y + z = 20

If fully occupied, x tables seating 4 will be occupied by 4x customers, y tables seating 6 will be occupied by 6y customers and z tables seating 8 will be occupied by 8z customers. Since we are told that if they occupy the tables with their seating, we will have a total of 108 customers. This gives the equation

4x + 6y + 8z = 108.

From th last statement "if only half ..." We can form the following equation:

(4x)/2 + (6y)/2 + (8z)/4 = 46.

From these we have the equations (after simplifying) to be:

x + y + z = 20 1

2x + 3y + 4z = 54 2

2x + 3y + 2z = 46 3

In order to solve these equations, let's subtract eqn 3 from 2, we have:

2z = 8

z = 4

Putting z = 4 in eqn 1 and eqn 2

We have:

x + y = 16 4

2x + 3y = 38 5

To solve these eqns, multiply eqn 4 by 2 and eqn 5 by 1.

4 x 2 : 2x + 2y = 32 6

5 x 1 : 2x + 3y = 38 7

Let's subtract eqn 6 from eqn 7, we have:

y = 6. Substituting y = 6 and z = 4 in eqn 1, we have x = 10.

Therefore, x = 10, y = 6 and z = 4