The problem has been reversed on me and I can't figure it out?

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Wow, someone who's actually trying to learn instread of just get answers to a problem set. Good on you.

Duke got the right answer but used two equations to do it. Your (Michelle)'s approach in your second problem is correct. Here's the same approach for your first problem:

x= pounds of 9% alloy

pounds of tin = 30(.11)=.09(x)+.15(30-x) Because the amount of 15% alloy must be 30 (total weight) - x (weight of 9%)=(30-x)
3.3 = 4.5 - .15x + .09x
-1.2 = -.06x
20 = x = pounds of 9% alloy
So 30-20=10 = pounds of 15% alloy.

Again, your approach works. Duke's approach works.

Your's is easier for this problem.

But you should get familiar with his approach of substituting an equation for one variable into a second equation. Not for stuff like this but for more complicated problems.

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There are 2 equations here. One is a tin-only equation and one is a total weight equation.

The amount of tin in the final alloy is (30)(.11)
the amount of tin in the first alloy is F(.15)
the amount of tin in the second alloy is S(.09)
So, the first equation is:
(30)(.11)=F(.15)+S(.09)

The second equation is the total weight of metal which is:
30=F+S where F is the weight of the first alloy and S is the weight of the second alloy.
Rearranging the second equation gives an equation for F:
F=30-S
Substituting the right side of this equation for F in the first equation gives:
(30)(.11)=(.15)(30)-(.15)S + (.09)S
Solving for S gives S=20
The weight of the first alloy is 30-20=10...Show more