A store sells notebooks for 3 each and pens for 2 each. A customer buys a combination of notebooks and pens totaling 10 items for $24. How…

Let $n$ be notebooks and $p$ pens. Then $n+p=10$ and $3n+2p=24$. Substitute $p=10-n$: $3n+2(10-n)=24$, so $n+20=24$ and $n=4$, choice H (and $p=6$). Check the cost: $3(4)+2(6)=12+12=24$. Choice F ($n=2$) gives $p=8$ and a cost of $6+16=22$. Choice G ($n=6$) gives $p=4$ and a cost of $18+8=26$, which is what you get by swapping the two counts. Choice J ($n=8$) gives $p=2$ and a cost of $28$. Only $n=4$ satisfies both the count and the total cost.