College Algebra

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A gift store is making a mixture of almonds, pecans, and peanuts, which sells for $4.50 per pound, $3 per pound, and $4 per pound, respectively. The storekeeper wants to make 50 pounds of mix to sell at $4.38 per pound. The number of pounds of peanuts is to be three times the number of pounds of pecans. Find the number of pounds of each to be used in the mixture.

work so far:
x = almonds
y = pecans
z = peanuts

x = $4.50
y = $3
z = $4

Thank you.

  • College Algebra -

    you don't seem to be reading what is given. If you set up your x,y,z as you did, then

    x+y+z = 50
    y = 3z
    4.50x + 3y + 4z = 4.38*50

  • College Algebra -

    To eliminate variables could I multiply the first equation by -4.50 and then add that equation to the third equation to eliminate x?

    Thank you.

  • College Algebra -

    This one is easier to solve just using substitution:

    y=3z, so

    x+3z+z = 50
    4.50x + 3(3z) + 4z = 4.38*50

    Simplify those to get

    x + 4z = 50
    4.50x + 13z = 219

    Now, since x = 50-4z,
    4.50(50-4z)+13z = 219
    225-18z+13z = 219
    5z = 6

    z = 1.2
    y=3z = 3.6
    x=50-4z = 45.2

    This makes sense, since the final price is so close to the price of peanuts. Most of the mix will have to be peanuts.

  • College Algebra -

    If you want to use elimination, arrange the equations so that x,y,z occur in order, and enter the coefficients into the array here:

    http://www.gregthatcher.com/Mathematics/GaussJordan.aspx

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