Respuesta :
so.. we'll use the decimal format of their percentage, thus, 11% is just 11/100 or 0.11, 15% is 15/100 or 0.15 and so on
so hmm [tex]\bf \begin{array}{lccclll} &amount&concentration& \begin{array}{llll} concentrated\\ amount \end{array}&\\ &-----&-------&-------\\ \textit{15\% alloy}&x&0.15&0.15x\\ \textit{9\% alloy}&y&0.09&0.09y\\ -----&-----&-------&-------\\ mixture&30&0.11&3.3 \end{array}[/tex]
whatever the amounts of "x" and "y" are, we know, their sum is 30 lbs, since that's what the 11% mixture weights
thus x + y = 30
now, the concentration of tin in each, must also add up to 3.3
thus 0.15x + 0.09y = 3.3
thus [tex]\bf \begin{cases} x+y=30\implies \boxed{y}=30-x\\ 0.15x+0.09y=3.3\\ ----------\\ 0.15x+0.09\left( \boxed{30-x} \right)=3.3 \end{cases}[/tex]
solve for "x", to see how much 15% alloy was used
what about "y"? well, y = 30 - x
so hmm [tex]\bf \begin{array}{lccclll} &amount&concentration& \begin{array}{llll} concentrated\\ amount \end{array}&\\ &-----&-------&-------\\ \textit{15\% alloy}&x&0.15&0.15x\\ \textit{9\% alloy}&y&0.09&0.09y\\ -----&-----&-------&-------\\ mixture&30&0.11&3.3 \end{array}[/tex]
whatever the amounts of "x" and "y" are, we know, their sum is 30 lbs, since that's what the 11% mixture weights
thus x + y = 30
now, the concentration of tin in each, must also add up to 3.3
thus 0.15x + 0.09y = 3.3
thus [tex]\bf \begin{cases} x+y=30\implies \boxed{y}=30-x\\ 0.15x+0.09y=3.3\\ ----------\\ 0.15x+0.09\left( \boxed{30-x} \right)=3.3 \end{cases}[/tex]
solve for "x", to see how much 15% alloy was used
what about "y"? well, y = 30 - x
