Answer:
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
