Question 1127524: Adrian, Nathan, and Brian bought lunch from the concession stand. Adrian paid $5.50 for burgers, fries and a drink; Nathan paid $4.00 for a burger and a drink; and Brian paid $4.25 for a burger and fries.
a) write three equations modeling the cost of their lunches using b, f, and d to represent the cost of a burger, fries, and a drink.
b)solve the system of equations to determine the cost of each food item.
i just need help setting it up
Found 3 solutions by addingup, ikleyn, MathTherapy:
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! Let burgers be x, fries y, and drink z
solving by elimination:
x + y + z = 5.50 = 11/2 multiply all times 2 to get rid of the fraction
x + 0 + z = 4.00 = 4
x + y + 0 = 4.25 = 17/4 multiply all times 4 to get rid of the fraction
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2x + 2y + 2z = 11
x + 0y + z = 4
4x + 4y + 0z = 17
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Swap rows 1 and 2:
x + 0y + z = 4
2x + 2y + 2z = 11
4x + 4y + 0z = 17
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Multiply the first equation times -2 and add to the 2nd equation:
-2x + -0y + -2z = -8
+2x + 2y + 2z = 11
---------------------
0x + 2y + 0z = 3
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
now we have:
x + z = 4
2y = 3
4x + 4y = 17
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
multiply the firs equation times -4 and add to the 3rd equation:
-4x + 0y -4z = -16
+
4x + 4y + 0z = 17
---------------------
0x + 4y + -4z = 1
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
now we have:
x + z = 4
2y = 3
4y - 4z = 1
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
swap rows 2 and 3:
x + z = 4
4y - 4z = 1
2y = 3; y = 3/2
Now that we know the value of y, let's find z:
4y - 4z = 1
4(3/2) - 4z = 1
6-4z = 1
-4z = -5
z = 5/4
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Now in the first equation solve for x by substituting y= 3/2 and z = 5/4. You can do this yourself, you should get 11/4 so that:
burger = 11/4 = $2.75
fries = 3/2 = $1.50
drink = 5/4 = $1.25
:
P.s. you can also solve this problem with a system of matrices using Cramer's Rule but you didn't say.
:
Happy learning!
Answer by ikleyn(52776)  (Show Source):
You can put this solution on YOUR website! .
I will show you MUCH FASTER way to solve the system !
The setup is still the same:
Let burgers be x, fries y, and drink z
x + y + z = 5.50 (1)
x + 0 + z = 4.00 (2)
x + y + 0 = 4.25 (3)
a) To find "y", subtract eq(2) from eq(1). You will get
y = 5.50 - 4.00 = 1.50. The unknown "y" is just found !
b) To find "z", subtract eq(3) from eq(1). You will get
z = 5.50 - 4.25 = 1.25. The unknown "z" is just found !
c) To find "x", substitute the found values of y and z into eq(1) to get
x + 1.50 + 1.25 = 5.50, which implies
x = 5.50 - 1.50 - 1.25 = 2.75.
Just solved (!)
The structure of the system (1), (2), (3) is such that elimination works very straightforward for "y" and "z" (!).
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An educated student must see it after a quick look into the system.
An experienced student must recognize this way of solving immediately after reading the condition even without writing the system.
This problem is, actually, for MENTAL solution.
Answer by MathTherapy(10551)  (Show Source):
You can put this solution on YOUR website! Adrian, Nathan, and Brian bought lunch from the concession stand. Adrian paid $5.50 for burgers, fries and a drink; Nathan paid $4.00 for a burger and a drink; and Brian paid $4.25 for a burger and fries.
a) write three equations modeling the cost of their lunches using b, f, and d to represent the cost of a burger, fries, and a drink.
b)solve the system of equations to determine the cost of each food item.
i just need help setting it up
Setup is as follows! The cost of a burger, an order of fries, and a drink cost are: b, f, and d, respectively
We then get: b + f + d = 5.5 ------ eq (i)
b + d = 4 -------- eq (ii)
b + f = 4.25 ----- eq (iii)
You can easily solve for each variable, since all have 1 as their coefficient. DO not follow that person who advised you to
do a million things and take an awfully long time to solve a SIMPLE, SIMPLE system in 3 equations. He's one of a few who'd make a very
simple problem EXTREMELY TIME-CONSUMING and COMPLEX. There is ABSOLUTELY no reason for someone to advise a person to do a problem that way.
Again, DON'T FOLLOW this person unless of course you like to torment yourself!!
b + f + d = 5.5 ------ eq (i)
b + d = 4 -------- eq (ii)
b + f = 4.25 ----- eq (iii)
f, or price of an order of fries =  ------ Subtracting eq (ii) from eq (i)
Now, LOOK how easy that was!
b + 1.5 = 4.25 ------ Substituting 1.5 for f in eq (iii)
b, or cost of burger = 
2.75 + 1.50 + d = 5.5 ------- Substituting 1.5 for f, and 2.75 for b in eq (i)
d, or cost of a drink = 
I went too far. You just wanted the setup but I went all the way! What the heck!
Maybe my solution will give you some insight into the easiest and most time-consuming way to do the problem.
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