Chocolate used at $0.90 per pound is 20 pounds #### Explanation: Let the weight of chocolates worth$0.90 a pound be ${c}_{0.90}$
Let the weight of chocolates worth $1.50 a pound be ${c}_{1.50}$Initial conditions: c_(0.90)+c_(1.50)=30^("lb")" ".....................Equation(1) The value of the blend is ($1.10)/("pound") so the total value of the final blend is ($1.10)/cancel("pound")xx30 color(white)("d")cancel("pound") =$33.00

Thus we have:

$0.90c_(0.90)+$1.5c_(1.5)=$33.00" "................Equation(2) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ You may choose which one you so wish to substitute for. I choose ${c}_{0.90}$Consider $E q n \left(1\right)$Write as ${c}_{0.90} = 30 - {c}_{1.50} \text{ } \ldots \ldots E q u a t i o n \left({1}_{a}\right)$Using $E q n \left({1}_{a}\right)$substitute for $\textcolor{red}{{c}_{0.90}}$in $E q n \left(2\right)$giving: Dropping the units of measerment for now. $\textcolor{g r e e n}{0.90 \textcolor{red}{{c}_{0.90}} \textcolor{w h i t e}{\text{dddd.d}} + 1.5 {c}_{1.50} = 33.00}$$\textcolor{g r e e n}{0.90 \left(\textcolor{red}{30 - {c}_{1.50}}\right) + 1.5 {c}_{1.50} = 33.00}$$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{d")27-0.9c_(1.50)color(white)("d}} + 1.5 {c}_{1.50} = 33.00}$$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{d")27color(white)("ddddd")+0.6c_(1.50)color(white)("dddd}} = 33.00}$Subtract 26 from both sides color(green)(0.6c_(1.50)=6 Divide both sides by $0.6$${c}_{1.50} = 10 \leftarrow \text{ pounds}$${c}_{0.90} = 30 - 10 = 20 \text{ pounds}\$