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How do you write a system of two linear equations from a real-world situation and interpret the solution in context?

Write systems of two linear equations given a table, a graph, a verbal description, or a real-world problem, then solve and interpret the solution in context (TEKS A.2H, A.2I).

A STAAR Algebra I answer on writing and modeling with systems of two linear equations from real-world situations (TEKS A.2H, A.2I), defining variables, building one equation per condition, and interpreting the solution.

Generated by Claude Opus 4.810 min answer

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  1. What this topic is asking
  2. Define variables, then one equation per condition
  3. Solving and interpreting
  4. Break-even and comparison problems
  5. How STAAR examines this topic
  6. Recognizing the common system patterns
  7. Try this

What this topic is asking

Modeling with systems is where Reporting Category 3 brings everything together. TEKS A.2H and A.2I ask you to write a system of two linear equations from a table, graph, description, or real-world problem, then solve it and interpret the solution. STAAR scores these on the modeling: defining variables clearly and building one correct equation per condition, not just producing a number.

Define variables, then one equation per condition

Two habits earn the modeling credit. First, define the variables with let statements: "let aa = adult tickets, cc = child tickets". Second, write one equation for each piece of information. Most STAAR systems follow a recognizable pattern:

  • A count equation: the number of items adds to a total, like a+c=12a + c = 12.
  • A value equation: each quantity times its unit value sums to a total, like 9a+5c=849a + 5c = 84.

Attaching the right number to the right variable, the price of an adult ticket to aa, is exactly where the distractors target.

Solving and interpreting

Once the system is written, solve by the most convenient method, then translate the answer back into the situation with units.

Break-even and comparison problems

A frequent application sets two expressions equal. In a break-even problem, cost (fixed plus per-unit) equals revenue (price times quantity), and the solution is the quantity where neither profit nor loss occurs. In a comparison problem (which plan is cheaper), set the two cost functions equal to find the crossover point, then reason about which side is cheaper.

How STAAR examines this topic

  • Multiple choice. Choose the system that models a situation; wrong-price-to-variable and swapped-total distractors are standard.
  • Equation editor and number entry. Build the equations and report the solution or the break-even quantity.
  • Multi-step items. Define variables, write the system, solve, and interpret, with credit spread across the steps.

A clarifying idea is that each equation captures a different constraint, so the solution is the unique combination satisfying both at once, which is why a real-world system needs exactly two independent conditions to pin down two unknowns.

Recognizing the common system patterns

Most STAAR system word problems fall into a few recognizable shapes, and naming the shape tells you what the two equations should be. A ticket or purchase problem pairs a count equation with a total-cost equation. A mixture problem (nuts, solutions, coins) pairs a total-amount equation with a total-value equation. A break-even or comparison problem pairs two cost-or-revenue expressions and sets them equal. A distance or age problem encodes two relationships between the unknowns. Spotting which pattern a question fits turns an intimidating paragraph into a familiar template.

The danger with the value equation is mismatched units: the count equation adds quantities (tickets, pounds, items), while the value equation adds money (price times quantity). Mixing them, such as adding a price into the count equation, produces a system that looks plausible but models the wrong situation. Keeping the two equations in their separate units, one counting and one valuing, is the discipline that makes the model correct and the interpretation meaningful.

Try this

Q1. Pens cost 2,notebooks2, notebooks 5. A student buys 8 items for $25. Write the system. [2 points]

  • Cue. p+n=8p + n = 8 and 2p+5n=252p + 5n = 25.

Q2. Cost C=5x+200C = 5x + 200, revenue R=9xR = 9x. Find break-even. [2 points]

  • Cue. 9x=5x+200⇒4x=200⇒x=509x = 5x + 200 \Rightarrow 4x = 200 \Rightarrow x = 50.

Exam-style practice questions

Practice questions written in the style of TEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

STAAR (style)1 marksMultiple choice. Adult tickets cost 9andchildticketscost9 and child tickets cost 5. A group buys 12 tickets for 84.Whichsystemmodelsthis?(A)84. Which system models this? (A) a + c = 12and and 9a + 5c = 84(B) (B) a + c = 84and and 9a + 5c = 12(C) (C) 9a + 9c = 12and and 5a + 5c = 84(D) (D) a + c = 12and and 5a + 9c = 84$
Show worked answer →

The correct answer is (A).

Let aa be adult tickets and cc be child tickets. The count equation is a+c=12a + c = 12. The cost equation multiplies each price by its quantity: 9a+5c=849a + 5c = 84. Choice (B) swaps the totals, choice (D) attaches the wrong price to each variable. One equation counts items; the other totals their value.

STAAR (style)3 marksEquation editor. A small business has start-up costs of 300andspends300 and spends 4 to make each item, while selling each item for $10. Write the cost and revenue equations and find the break-even number of items.
Show worked answer →

Cost C=4x+300C = 4x + 300, revenue R=10xR = 10x, and the business breaks even at x=50x = 50 items.

Cost combines the fixed start-up and the per-item cost: C=4x+300C = 4x + 300. Revenue is price times quantity: R=10xR = 10x. Break-even is where cost equals revenue: 10x=4x+30010x = 4x + 300, so 6x=3006x = 300 and x=50x = 50. At 50 items, both cost and revenue are $500. Below 50 the business loses money; above 50 it profits. Defining the variables and writing one equation per quantity earns the modeling credit.

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