Question

A metallugarist has an alloy with 6% titanium and an alloy with 20% titanium. He needs 100 grams of an alloy with 13% titanium. How much of each alloy should be mixed to attain the 100 grams of alloy with 13% titanium?

Answer 1

`\bf \begin{array}{lccclllll} &amount(g)&\textit{\% of titanium}&\textit{grams of titanium}\\ &-----&-----&-------\\ \textit{6\% alloy}&x&0.06&0.06\cdot x\\ \textit{20\% alloy}&y&0.2&0.2\cdot y\\ ----&-----&-----&-----\\ mixture&100&0.13&100\cdot 0.13\to 13 \end{array}`

so.. .whatever those amounts are, of "x" and "y", for the mixture,
they must add up to 100 grams, or x + y = 100

and the titanium percentage of that sum, must yield a 13 grams of titanium on those 100 grams

thus
`\bf \begin{cases} x+y=100\to y=\boxed{100-x} \\\\ 0.06x+0.2\boxed{y}=13 \end{cases}`

do the substitution, and solve for "x",

what's "y" amount? well, y = 100 - x