Question

I've tried this for an Hour, and can't get it.

A metallurgist has an alloy with 15% TITANIUM and an alloy with 30% TITANIUM. He needs 100 GRAMS of an alloy with 21% TITANIUM. How much of each alloy should be mixed to attain the 100 Grams of Alloy with 21% Titanium?

Answer 1

I know i need 2 equations and I have to use elimation to find the answers.

Answer 2

x= amount of 15% alloy
100-x= amount of 30% alloy to use
.15x= amount of titanium in x grams of 15% alloy
.30(100-x)= amount of titanium in 100-x grams of 30 % alloy
.21(100) = amount of titanium in 100 grams of 21% alloy
.15x+.30(100-x)=.21(100)
Or in other words, initial titanium + added titanium = final titanium

Answer 3

\[.15x+.3(100-x)=.21(100)\]
\[.15x+30-.3x=21\]
\[-.15x=-9\]
\[x=60\]
Use 60 g of15% alloy and 40 g of 30% alloy
And NO. You don't need two equations. Although you could do it with 2 and they would be:
x+y=100
.15x+.3y=.21(100)

Answer 4

Thanks, I was setting it up wrong at the begining. I understand now. Thank you

Answer 5

yw