SOLUTION: A metallurgist made two purchases. The first purchase which cost $1080 included 30kilograms of iron and 45 kilograms of lead. The second purchase at the same prices cost $372 and i

Question 1022198: A metallurgist made two purchases. The first purchase which cost $1080 included 30kilograms of iron and 45 kilograms of lead. The second purchase at the same prices cost $372 and included 15 kilograms of iron and 12 kilograms of lead. Find the cost per kilogram of the iron and lead.
Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
30i+45l=1080
15i+12l= 372 Multiply both sides of this equation times 2 and subtract from the equation above, like this:
30i+45l / 1080
-
30i+24l = 744
----------------
...21l = 336 divide both sides by 21:
l = 16 So now we know that lead is 16. Let's find iron by plugging lead into one of the equations, any one of the two. I'll do the second equation:
15i+12(16) = 372
15i+192 = 372 deduct 192, both sides
15i = 180 divide both sides by 15
i = 12
Do the other one, I'll help you:
30i+45l = 1080 plug in the value for i and for l, the result has to be 1080 to be right:
30(12)+45(16)= 1080
360+720 = 1080 Add on left and you'll find that we have the right answer.

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