6.6 Creating Equations, Inequalities, & Constraint Models
Key Takeaways
- Creating an equation in one variable starts by naming the unknown in words, then translating each sentence of the problem into one algebraic statement.
- Creating an equation in two variables produces a graph, and the axes must be labeled with the quantity and its unit and scaled to the data.
- Constraints in a modeling problem become inequalities or a system of inequalities, and the solution region is the overlap of all of them.
- A mathematically correct solution can still be non-viable in context — negative hours, fractional people, or a value beyond a stated budget must be rejected.
Creating Equations, Inequalities, & Constraint Models
Solving an equation is mechanical. Writing the equation is where the points are, and Level A tests it as a distinct skill across three standards: create equations and inequalities in one variable; create equations in two or more variables and graph them "on coordinate axes with labels and scales"; and represent constraints by equations, inequalities, or systems, "interpreting solutions as viable or non-viable options in a modeling context."
Part 1: One-Variable Equations and Inequalities
The reliable procedure is four steps, and skipping step 1 is what causes most errors.
- Define the variable in words. "Let $h$ = the number of hours of labor." Not just "let x = hours" — write the unit.
- Translate one sentence at a time.
- Solve.
- Answer the question that was asked, which is often not the variable you solved for.
Problem. A repair shop charges a $65 diagnostic fee plus $95 per hour. A customer's total bill was $445. How long did the repair take?
| Sentence | Algebra |
|---|---|
| Let $h$ = hours of labor | $h$ |
| $95 per hour | $95h$ |
| plus a $65 fee | $95h + 65$ |
| total bill was $445 | $95h + 65 = 445$ |
The same problem as an inequality
The customer can spend at most $445. What is the greatest number of hours she can afford?
Translate the boundary language carefully — this is a scored distinction:
| Phrase | Symbol |
|---|---|
| at most, no more than, maximum, up to | $\le$ |
| at least, no less than, minimum | $\ge$ |
| more than, exceeds, over | $>$ |
| less than, under, below | $<$ |
Part 2: Two-Variable Equations and Their Graphs
When two quantities vary together, the model is a two-variable equation and the answer is a graph, not a number.
A landscaping crew charges $40 per lawn and $25 per hedge. Write an equation for the revenue $R$ from $\ell$ lawns and $h$ hedges.
If total revenue is fixed at $400, the equation $40\ell + 25h = 400$ describes every combination that produces that revenue, and its graph is a line segment.
Graphing with labels and scales
The standard explicitly requires labels and scales, and TABE items ask about them:
- Label each axis with the quantity and its unit — "Lawns mowed," "Revenue (dollars)" — not just "x" and "y."
- Choose a scale that fits the data. If revenue runs to $400, tick every $50, not every $1.
- Start at a sensible origin. For counts of jobs, both axes begin at 0.
- Plot the intercepts first. Setting $h = 0$ gives $\ell = 10$; setting $\ell = 0$ gives $h = 16$. Connect $(10, 0)$ and $(0, 16)$.
Part 3: Constraints and Systems
A constraint is a limit the situation imposes. Each one becomes an inequality, and together they carve out a region of acceptable solutions.
Problem. A food-truck owner makes tacos and burritos. Each taco uses 2 oz of beef and each burrito uses 5 oz. She has 200 oz of beef. She must make at least 20 tacos, and she cannot make a negative number of anything.
| Constraint in words | Inequality |
|---|---|
| Beef supply is limited to 200 oz | $2t + 5b \le 200$ |
| At least 20 tacos | $t \ge 20$ |
| Cannot make negative burritos | $b \ge 0$ |
The set of $(t, b)$ pairs satisfying all three at once is the feasible region. Any point inside it is a workable production plan; any point outside violates something.
Testing a candidate plan. Is 30 tacos and 30 burritos feasible?
- $2(30) + 5(30) = 60 + 150 = 210$, and $210 \le 200$ is false.
The plan fails on beef supply, so it is rejected — even though $t \ge 20$ and $b \ge 0$ both hold. All constraints must be satisfied simultaneously.
Part 4: Viable vs. Non-Viable Solutions
This is the modeling judgment TABE actually asks about, and it is the difference between an algebra answer and a correct answer.
| Algebraic result | Context | Viable? |
|---|---|---|
| $t = 4.5$ hours of labor | billed in quarter hours | yes |
| $b = 12.7$ buses needed | buses are whole | no — round up to 13 |
| $n = -3$ employees | headcount | no — reject; check the setup |
| $p = $0$ price | must cover a $5 cost | no — outside the stated constraint |
| $x = 250$ units | machine capacity is 200 | no — violates a constraint |
Worked viability check. A ticket equation gives $a = 8.4$ adult tickets and $c = 11.6$ child tickets. Tickets are whole objects, so this exact solution is non-viable. The correct response is to report that no whole-number solution exists for the stated totals, or to identify the nearest whole-number pair that satisfies the constraints — not to round silently and present 8 and 12 as if they solved the original system.
The habit to build: after every modeling answer, ask two questions. Can this quantity be negative? Can it be a fraction? If the answer to either is no and your solution says otherwise, the model or the arithmetic needs another look.
A moving company charges a flat $120 truck fee plus $55 per hour. A customer has budgeted no more than $505. Which inequality models the situation, and what is the greatest whole number of hours she can book?
A community garden has 180 feet of fencing for a rectangular plot and must be at least 30 feet long. Which system of constraints correctly models the situation?
A workshop needs to seat 94 participants. Each table seats 8. An organizer solves 8t = 94 and gets t = 11.75. What is the correct modeling response?