Question 164616
let x = amount of grams in first alloy.
let y = amount of grams in second alloy.
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question is how many grams of each alloy is needed to make 60 grams of gold and 88 grams of lead.
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number of grams of gold in first alloy is .10 * x.
number of grams of gold in second alloy is .30 * y.
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number of grams of lead in first alloy is .15 * x.
number of grams of lead in second alloy is .4 * y.
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total number of grams of gold in third alloy is 60.
total number of grams of lead in third alloy is 88.
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total grams of gold from first and second alloy must = 60.
equation looks like.
.10*x + .30*y = 60
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total grams of lead from first and second alloy must = 88.
equation looks like.
.15*x + .4*y = 88
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solve these equations simultaneously.
.10*x + .30*y = 60 (first equation)
.15*x + .4*y = 88 (second equation)
multiply both sides of first equation by 1.5
.15*x + .45*y = 90
subtract first equation from second eqution.
.05*y = 2
divide both sides by .05
y = 40
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substitute 40 for y in first equation.
.10*x + .30*(40) = 60
.10*x + 12 = 60
.10*x = 48
x = 480
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substitute 40 for y and 480 for x in second equation.
.15*480 + .45*40 = 90
72 + 18 = 90
90 = 90
x = 480 and y = 40 looks good in second equation.
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answer is you need 480 grams of first alloy and 40 grams of second alloy.
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