Respuesta :
so, let's say, "p" is how many lbs of peanuts, "a" of almonds and "r" of raisins
[tex]\bf \begin{array}{lccclll} &amount&price&\textit{amount price}\\ &-----&-----&-------\\ peanuts&p&1.5&1.5p\\ almonds&a&3.00&3a\\ raisins&r&1.5&1.5r\\ -----&-----&-----&-------\\ mixture&11&&21 \end{array}[/tex]
so.... whatever p,a,r are, they will sum up to 11, thus
p+a+r = 11
and whatever 1.5p+3a+1.5r is, must add up to 21
1.5p+3a+1.5r = 21
now.. .we know that the mixture uses twice as many peanuts as almonds, namely, if there's "a" amount of almonds, then of peanuts is twice that much or 2*a, or 2a, thus p = 2a
thus [tex]\bf \begin{cases} p+a+r=11\qquad \qquad p=2a\\ 2a+a+r=11\\ 3a+r=11\\ \boxed{r}=11-3a\\ ----------\\ 1.5p+3a+1.5r=21\\ 1.5(2a)+3a+1.5r=21\\ 3a+3a+1.5r=21\\ 6a+1.5r=21\\ 6a+1.5(\boxed{11-3a})=21 \end{cases}[/tex]
solve for "a" to see how many lbs of almonds will be needed
what about raisins? well, r = 11 - 3a
and peanuts? well, p = 2a
[tex]\bf \begin{array}{lccclll} &amount&price&\textit{amount price}\\ &-----&-----&-------\\ peanuts&p&1.5&1.5p\\ almonds&a&3.00&3a\\ raisins&r&1.5&1.5r\\ -----&-----&-----&-------\\ mixture&11&&21 \end{array}[/tex]
so.... whatever p,a,r are, they will sum up to 11, thus
p+a+r = 11
and whatever 1.5p+3a+1.5r is, must add up to 21
1.5p+3a+1.5r = 21
now.. .we know that the mixture uses twice as many peanuts as almonds, namely, if there's "a" amount of almonds, then of peanuts is twice that much or 2*a, or 2a, thus p = 2a
thus [tex]\bf \begin{cases} p+a+r=11\qquad \qquad p=2a\\ 2a+a+r=11\\ 3a+r=11\\ \boxed{r}=11-3a\\ ----------\\ 1.5p+3a+1.5r=21\\ 1.5(2a)+3a+1.5r=21\\ 3a+3a+1.5r=21\\ 6a+1.5r=21\\ 6a+1.5(\boxed{11-3a})=21 \end{cases}[/tex]
solve for "a" to see how many lbs of almonds will be needed
what about raisins? well, r = 11 - 3a
and peanuts? well, p = 2a