SAT Question ee7b1de1: Linear Inequalities answer and explanation with Desmos solution

Let $s$ represent the number of small candles the owner can purchase, and let $l$ represent the number of large candles the owner can purchase. It's given that the owner pays $$4.90$ per candle to purchase small candles and $$11.60$ per candle to purchase large candles. Therefore, the owner pays $4.90s$ dollars for $s$ small candles and $11.60l$ dollars for $l$ large candles, which means the owner pays a total of $4.90s+11.60l$ dollars to purchase candles. It's given that the owner budgets $$2,200$ to purchase candles. Therefore, $4.90s+11.60l\le 2,200$. It's also given that the owner must purchase a minimum of $200$ candles. Therefore, $s+l\ge 200$. The inequalities $4.90s+11.60l\le 2,200$ and $s+l\ge 200$ can be combined into one compound inequality by rewriting the second inequality so that its left-hand side is equivalent to the left-hand side of the first inequality. Subtracting $l$ from both sides of the inequality $s+l\ge 200$ yields $s\ge 200-l$. Multiplying both sides of this inequality by $4.90$ yields $4.90s\ge 4.90\left(200-l\right)$, or $4.90s\ge 980-4.90l$. Adding $11.60l$ to both sides of this inequality yields $4.90s+11.60l\ge 980-4.90l+11.60l$, or $4.90s+11.60l\ge 980+6.70l$. This inequality can be combined with the inequality $4.90s+11.60l\le 2,200$, which yields the compound inequality $980+6.70l\le 4.90s+11.60l\le 2,200$. It follows that $980+6.70l\le 2,200$. Subtracting $980$ from both sides of this inequality yields $6.70l\le 2,200$. Dividing both sides of this inequality by $6.70$ yields approximately $l\le 182.09$. Since the number of large candles the owner purchases must be a whole number, the maximum number of large candles the owner can purchase is the largest whole number less than $182.09$, which is $182$.