Ajeweler maker has 20 ounces of a gold alloy worth $160 an ounce and 10 ounce of a silver alloy worth $140 an ounce. how many ounces of another alloy at $124 an ounce must she add to these in order to have a mixture of the three that she can use to make 3-ounce jewelry worth $438 a piece?

For this case he does not need anything of $ 124 / oz. Let's prove it.
 Let
 x = gold alloy required
 (3-x) = silver alloy required
 160x + 140 (3-x) = 438
 160x + 420 - 140x = 438
 160x - 140x = 438 - 420
 20x = 18
 x = 18/20
 x = 0.9 oz gold alloy
 3 - 0.9 = 2.1 oz silver alloy
 We check the result:
 0.9 (160) + 2.1 (140) =
 144 + 294 = 438
 
 No ounces of another alloy at $ 124 an ounce must she add to these in order to have a mixture of the three that she can use to make 3-ounce jewelry worth $ 438 a piece

1. isolate the variable, in this case, d. subtract 35.5 on both sides of the equation.2. 52 subtracted by 35.5 is 16.53. therefore your answer would be d= 16.5this is the basics of pre algebra, isolating variable through adding, subtracting, multiplying, and dividing, if you have any other questions to ask just comment!