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Math Help - Systems of Equations Story Problem

  1. #1
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    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?
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  2. #2
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    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.
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