SAT Question f224df07: Linear Inequalities answer and explanation with Desmos solution

Choice C is correct. Let $a$ equal the number of $120$-pound packages, and let $b$ equal the number of $100$-pound packages. It's given that the total weight of the packages can be at most $1,100$ pounds: the inequality $120a+100b\le 1,100$ represents this situation. It's also given that the helicopter must carry at least $10$ packages: the inequality $a+b\ge 10$ represents this situation. Values of $a$ and $b$ that satisfy these two inequalities represent the allowable numbers of $120$-pound packages and $100$-pound packages the helicopter can transport. To maximize the number of $120$-pound packages, $a$, in the helicopter, the number of $100$-pound packages, $b$, in the helicopter needs to be minimized. Expressing $b$ in terms of $a$ in the second inequality yields $b\ge 10-a$, so the minimum value of $b$ is equal to $10-a$. Substituting $10-a$ for $b$ in the first inequality results in $120a+100\times (10-a)\le 1,100$. Using the distributive property to rewrite this inequality yields $120a+1,000-100a\le 1,100$, or $20a+1,000\le 1,100$. Subtracting $1,000$ from both sides of this inequality yields $20a\le 100$. Dividing both sides of this inequality by $20$ results in $a\le 5$. This means that the maximum number of $120$-pound packages that the helicopter can carry per trip is $5$.