College Algebra

posted by Gabbi

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.

1. Steve

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

2. Gabbi

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.

3. Steve

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.

4. Steve

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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