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How many pounds of nuts worth $1.10 per pound must be mixed with 36 pounds of candy worth $1.60 per pound to produce a party mix worth $1.50 per pound?

User Bsisco
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\bf \begin{array}{lccclll} &amount&price&\textit{total price}\\ &-----&-----&-----\\ \textit{nuts \$1.10/lb}&x&1.10&1.10x\\ \textit{candy \$1.60/lb}&36&1.6&57.6\\ -----&-----&-----&-----\\ \textit{mixture \$1.50/lb}&y&1.50&1.50y \end{array}

so... whatever "x" may be, we know that the whole mix will be "y" amount, thus x + 36 = y

and their total prices will also add up to 1.50y
thus 1.10x + 57.6 = 1.50y

thus
\bf \begin{cases} x+36=\boxed{y}\\ 1.10x+57.6=1.50y\\ ----------\\ 1.10x+57.6=1.50\left( \boxed{x+36} \right) \end{cases}

solve for "x", to see how much nuts will be added

what about "y"? well x + 36 = y
User Iswar
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