Solve multi-step inequalities, including reversing the inequality when appropriate.
Model a multi-step context with an inequality
Problem
A \(\$12\) fee plus \(\$4~\text{per}~\text{student}\) must total at most \(\$40\). Let \(s\) be the nonnegative whole-number student count. Translate the cost constraint into an inequality, solve it over the contextual domain, and report the inequality with its complete solution set.
Big Picture
What this problem is really about
Build the total from a twelve-dollar fee that occurs once and a four-dollar charge repeated for each student. “At most” makes forty an included upper limit, so solve by removing the fixed fee and dividing by the positive rate. Then restrict the result to nonnegative whole-number student counts and include every feasible count, not only the greatest one.
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