# Thread: Systems of Equations Story Problem

1. ## Systems of Equations Story Problem

One pan pizza and two beef burritos provide 1980 calories. Two pan pizzas and one beef burrito provide 2670 calories. Use a system of equations to find the caloric content of each item. Define the variables:

x = pan pizza y = beef burrito

equation 1: x+2y=1980
equation 2: 2x+y=2670

How do I solve this?

2. Let x be the "caloric content" of 1 pan pizza; y be the "caloric content" of 1 beef burrito.

x + 2y = 1980 --(WTF)
2x + y = 2670 -- (WTH)

lol, you have to solve it simultenously, which goes this way:

One way: 2 times (WTF); you want to do this to eliminate the unknown x.

=> 2x + 4y = 3960 -- (2WTF)
~> 2x + y = 2670 -- (WTH)

(2WTF) - (WTH) = 2x - 2x + 4y - y = 0 + 3y = 3y = 3960 - 2670 = 1290

=> 3y = 1290
=> y = 430

Since " 2x+y=2670 ", sub " y = 430 " to that equation to get an expression of total calories in terms of x.

=> 2x + 430 = 2670
=> 2x = 2670 - 430 = 2240
=> x = 2240/2 = 1120

Alternatively, you can use that elimination method to get rid of y as well, leaving x to then solve for x.

Second way (by subsituation):

x+2y=1980
=> x = 1980 - 2y -- (WTF)
2x+y=2670 -- (WTH)

Sub (WTF) to (WTH)

(WTH) = 2(1980 - 2y) + y = 2670
=> 3960 - 4y + y = 2670
=> -3y = -1290
=> y = (-1290)/(-3) = 430 -------( which is equal to the y value I initially found using the first method )

Subsequently, you solve for the value of x.
You can do this alternatively, using the second method, by making y the subject, then blah blah.

Hehe, hope this helps.