Budget Planner

Pre-Algebra · Axiom Academy

LESSON Planning a Party on a Budget A budget is just an inequality in disguise: keep the running total at or below the limit, and solving it tells you exactly how much you can buy. You rent a hall for a flat 50 , then buy snack trays at 8 each . Your total has two parts: the one-time fee plus 8 added for every tray. If n is the number of trays, the cost is 50 + 8n . Your budget is 130 . The only rule is that the total can't exceed it — that "at most" is what makes this an inequality, not an equation. In the animation, each tray stacks a bar onto the running total, and the total climbs toward the red budget line. The first tray that pushes the bar past the line is the one you can't afford. Total cost = fixed fee + cost per tray × number of trays The budget rule: total stays at or below 130 2. Solving It: How Many Can You Afford? "How many trays fit the budget?" means: find every n that makes true. We solve it the same way we solve an equation — undo the operations on n , doing the same thing to both sides — with one rule to remember below. Take the 50 hall fee off both sides: 50 + 8n 130 becomes 8n 80 . What's left is the money available for trays. Each tray is 8 , so divide both sides by 8 : 8n 80 becomes n 10 . You can buy at most 10 trays. is a whole set of answers: all work. The inequality describes every affordable choice at once. If you ever multiply or divide both sides by a negative number, flip the inequality sign . (Here we divided by +8 , so it stays .)

This is the written version of the interactive lesson above. See the full Pre-Algebra course.