A large family wedding for the Smiths will require 113 trays of sandwiches, 116 trays of hors d' oeuvres, and 144 trays of sweets. Hands-Off caterers can make 15 trays of sandwiches, 10 trays of hors d'oeuvres, and 12 trays of sweets in one day. Hands-Free caterers can make 10 trays of sandwiches, 13 trays of hors d'oeuvres, and 14 trays of sweets in a day. Hand-Some caterers can make 12 trays of sandwiches, 15 trays of hors d'oeuvres, and 20 trays of sweets in a day. ****All three caterers are hired. How many days in advance of the wedding must each caterer begins their preperation of the food? Hint: 3 equations with 3 unknowns.
Btw, I know I need to create three equations and solve it using elimination. 1 for each caterer, x, y, and z.
oh and each will represent HOW MANY DAYS MUST EACH CATERING PARTY START TO PREPARE FOOD BEFORE THE WEDDING.
15x + 10y + 12z = 113
10x + 13y + 15z = 116
12x + 14y + 20z = 144
To solve the system of equations, you can use the method of elimination. Here are the steps to find the number of days each caterer should start preparing food:
Step 1: Write down the system of equations:
15x + 10y + 12z = 113
10x + 13y + 15z = 116
12x + 14y + 20z = 144
Step 2: Choose a variable to eliminate first. Let's eliminate the variable "x" by multiplying both sides of equation 1 by 10, and equation 2 by 15:
150x + 100y + 120z = 1130
150x + 195y + 225z = 1740
Step 3: Subtract equation 1 from equation 2 to eliminate "x":
0 + 95y + 105z = 610
Step 4: Now, we need to eliminate another variable. Let's eliminate "x" again, this time by multiplying equation 1 by 12, and equation 3 by 15:
180x + 120y + 144z = 1356
180x + 210y + 300z = 2160
Step 5: Subtract equation 1 from equation 3 to eliminate "x":
0 + 90y + 156z = 804
Step 6: We now have two equations in two variables. We can solve these equations simultaneously. Multiply the previous equation (Step 3) by -12, and the previous equation (Step 5) by 10:
-1140y - 1260z = -7320
900y + 1560z = 8040
Step 7: Add equation 6 to equation 7 to eliminate "y":
-240y + 300z = 720
Step 8: Divide equation 8 by 60:
-4y + 5z = 12
Step 9: Now, we can solve this equation for "y" in terms of "z":
-4y = -5z - 12
y = (5z + 12) / 4
Step 10: Substitute the value of "y" in terms of "z" into equation 6:
(-1140(5z + 12) / 4) - 1260z = -7320
Step 11: Simplify equation 10 and solve for "z":
-2850z - 6840 - 5040z = -7320
-7890z = -7320 + 6840
-7890z = -480
z = (-480) / (-7890)
z ≈ 0.0607
Step 12: Substitute the value of "z" into equation 9 to solve for "y":
y = (5z + 12) / 4
y ≈ (5(0.0607) + 12) / 4
y ≈ 12.152 / 4
y ≈ 3.038
Step 13: Substitute the values of "y" and "z" into equation 1 to solve for "x":
15x + 10y + 12z = 113
15x + 10(3.038) + 12(0.0607) = 113
15x + 30.38 + 0.7284 = 113
15x + 31.1084 = 113
15x ≈ 81.8916
x ≈ 5.459
So, the solution to the system of equations is approximately:
x = 5.459
y = 3.038
z = 0.0607
Therefore, Hands-Off caterers should start preparing 5.459 days before the wedding, Hands-Free caterers should start preparing 3.038 days before the wedding, and Hand-Some caterers should start preparing 0.0607 days before the wedding.
Since we can't start preparing food less than a day before the wedding, Hand-Some caterers should start their preparation on the day of the wedding.