Can someone help explain the steps to this problem?

Setup you equation like this, where B is the weight of the second alloy. Your total amount of tin stays constant so you setup the left side of the equation as the amount of tin before the alloys are combined, the right side is the amount of tin afterwards:

0.15(20)+0.10(B)=0.12(20+B)

That was the hard part, now it's just a standard 1 variable equation to solve:

reduce the equation:

3 +.1B = 2.4 +.12B

combine like terms:

0.6 = .02B

solve for B:

B = 30.0 lb

.15 * 20 + .10 *x = .12 * (20 + x)

basically, you have to set it up such that the total tin content is the same from what you're mixing (20 lbs of .15 tin and x pounds of .10 tin) to what you're making (20 plus x pounds of .12 tin)

3 + .1 x = .12 x + 2.4

.6 = .02 x

x = 30 lbs

Let x = No. of lbs of 2nd alloy to add

Total weight after mixture = x + 20

Form equation of tin

(0.12)(x+20) = (0.15)(20) + (0.10)x

0.12x + 2.4 = 3 + 0.10x

0.12x - 0.10x = 3.0 - 2.4

0.02x = 0.6

x = 0.6 / 0.02 = 30 lbs of 2nd alloy

working equation: 0.15x + 0.10y = 0.12(x + y)

You have an infinite # of solutions.

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