Word problems, constraints and checking solutions

Identify what must be found and what is given

An equation application begins with a question about a situation. Before calculating, restate what must be found: a number of tickets, a pair of side lengths or a completion time. Separate the supplied data from the unknowns. Keep words such as exactly, at least, remaining and together, because each changes what an acceptable answer must satisfy.

Suppose an event sells 21 tickets at two prices and records its total receipts. The number 21 is the ticket total, not the money total; the question might ask for the count at each price. A labelled sketch or a small table can organise this information. Do not infer an extra discount, fee or ticket type unless the question supplies it. An equation should express the stated situation and any assumptions you explicitly choose.

Name each variable and its permitted values

Write a declaration before using a letter. For example, let a be the number of adult tickets and c the number of child tickets. Both a and c must be nonnegative integers: zero tickets of one type can be allowed, but a negative or fractional ticket count cannot. Distinguish this restriction from a positive length or a positive elapsed time, which need not be an integer.

When writing a numerical equation for measurements, specify the unit used for each value. If w is the numerical width in metres and L the numerical length in metres, the actual sides are w m and L m. The conditions are w > 0 and L > 0. If t is a completion time measured in hours, write t > 0. Record these domains before solving, so the algebraic answer is judged against the same problem throughout.

Translate the complete phrase and preserve its order

Build an equation from the relationship between named quantities. If a length L is 3 metres more than a width w, with both expressed in metres, write L = w + 3. If L is 3 metres less than twice w, write L = 2w − 3. The whole phrase determines which quantity is doubled and which is reduced. The words less than do not mean that you may subtract in any order.

Read your equation back into words. For L = 2w − 3, first double the width, then subtract 3 metres to obtain the length. This is different from L = 2(w − 3), which doubles the width after subtracting 3 metres. Similarly, the difference of two quantities must identify their order, and an at-least condition becomes an inequality. Choosing an operation from a single word such as total can hide whether the total is given or is what you must find.

Give each independent relationship its own equation

For adult tickets costing €8 and child tickets costing €5, the statement that exactly 21 tickets were sold gives a + c = 21. Receipts of exactly €135 give 8a + 5c = 135, with the amounts expressed in euros. The first equation counts tickets; the second adds their money contributions. Multiplying a ticket count by its price per ticket produces money, not another ticket count.

Both equations concern the same sale and must hold for the same pair (a, c). A solution of only one is not yet a solution of the system. You may instead use the single variable a and substitute c = 21 − a, giving 8a + 5(21 − a) = 135. Then the original count restrictions require a to be an integer from 0 to 21. This changes the representation of the model without changing its conditions.

Choose compatible units before substituting

An equality must compare the same kind of quantity on both sides. Add metres to metres or money to money; do not add a length directly to an area. For a rectangle with sides w m and (w + 3) m and area 40 m², the numerical area equation is w(w + 3) = 40 because the length values use metres and the area value uses square metres. If the given side difference were 300 cm, convert it to 3 m before forming this equation.

A rate also carries a unit. At a constant speed of 12 km/h for 35 minutes, first convert the time to 35/60 = 7/12 hour. Then distance is 12 × 7/12 = 7 km. Multiplying 12 by 35 would incorrectly treat minutes as hours. You could instead express the speed in kilometres per minute; either consistent choice gives the same distance. Check the unit as well as the number when returning to the question.

Choose a method from the simplified model

The situation determines the relationships; their algebraic form helps you choose a method already learned. A single linear equation can use equal operations on both sides. Two linear equations can use substitution or elimination. A positive-area rectangle with one side expressed using the other may produce a quadratic, solvable by factoring, completing the square or the quadratic formula. Collect and simplify terms before identifying the type.

A constant-rate work model can produce a rational equation. If two printers would finish the same divisible job alone in 6 hours and 9 hours, their rates are 1/6 and 1/9 of that job per hour. Assuming both begin together, keep their individual rates, do not duplicate work and have no pauses or shared bottleneck, the combined time t hours satisfies 1/6 + 1/9 = 1/t, with t > 0. Add the rates, not the completion times. For a model with variable denominators, keep every original denominator nonzero before clearing them; for square roots, retain the original real-domain and sign conditions.

Check each candidate in the original model and context

After solving, substitute every candidate into the original equations, not just a transformed equation. This matters especially after clearing denominators or squaring. An excluded denominator remains excluded even if a later expression cancels it. Squaring can also produce a candidate that fails the original sign condition. A successful substitution establishes that candidate; the solution method must also justify that every permitted possibility was considered.

Then test every context constraint, including those absent from the equations you solved. The ticket equations above have the unique real pair a = 10 and c = 11. The checks 10 + 11 = 21 and 8 × 10 + 5 × 11 = 135 both pass, and both counts are permitted. If the question also required at least 12 adult tickets, this same pair would fail that additional constraint. Because the original equations have no other real pair, the full constrained problem would then have no solution. Passing the equations alone does not establish context feasibility.

Rounding cannot repair an impossible exact count

Change the ticket receipts to exactly €136 while keeping 21 tickets and the two prices unchanged. Substituting c = 21 − a gives 8a + 5(21 − a) = 136, hence 3a = 31. The unique real pair is a = 31/3 and c = 32/3. It satisfies both numerical equations but neither value is an integer. Therefore the exact ticket problem has no solution in its permitted domain.

Rounding those counts to 10 and 11 preserves the ticket total but gives receipts 8 × 10 + 5 × 11 = 135, not 136. Rounding a measurement or a reported time can be appropriate when an approximation is requested; it cannot make a forbidden fractional count satisfy exact conditions. Keep fractions and radicals exact until the checks are complete, then state any requested approximation and its precision. Do not quietly replace exact given data with nearby values.

Determine whether the information fixes an answer

A rectangle's perimeter of 24 m gives 2x + 2y = 24, or x + y = 12, for positive side values x and y in metres. This information alone does not determine a unique rectangle. The pairs (5, 7) and (4, 8) both pass; more generally, every real x with 0 < x < 12 gives a permitted partner y = 12 − x. A further independent condition, such as an area or a side relationship, may narrow the possibilities.

Different wording can supply incompatible conditions rather than too few. If the same two ticket counts must satisfy both a + c = 21 and a + c = 22, subtracting the equations gives 0 = 1. No pair satisfies them together, even over the real numbers. Repeating a + c = 21 as 2a + 2c = 42 supplies no additional restriction. Decide whether the actual relationships are independent and compatible; the number of written equations alone does not guarantee a unique answer.

State the answer with its meaning and assumptions

Finish by answering what was asked, including the unit and all permitted alternatives. In the rectangle model w(w + 3) = 40, algebra gives w = 5 or w = −8. The positive-length domain admits only w = 5, so the sides are 5 m and 8 m. The rejected value does satisfy the numerical equation: (−8)(−5) = 40. It fails because those numbers cannot represent the specified positive side lengths, not because negative algebraic roots are always invalid.

A mathematically checked model gives a conclusion conditional on its data and assumptions. In a work problem, constant independent rates and freely divisible work must be stated; actual printers might pause, share a bottleneck or have setup time. An observed completion time could then differ without changing the algebra within the stated model. Distinguish given exact data, measured or rounded data and chosen assumptions. If the result is infeasible or the data do not determine one answer, say so instead of inventing an extra condition.

Compare equation solutions with context feasibility

The model offers six stated cases: the €135 ticket sale, the €136 sale with fractional algebraic counts, the rectangle with area 40 m², the two constant-rate printers, a rectangle whose only supplied condition is perimeter 24 m, and the €135 ticket sale with the extra requirement of at least 12 adult tickets. Choose a candidate independently of the case, then compare its original-equation checks with its domain and additional constraints.

A finite list of candidates is a way to inspect checks, not a description of every allowed real value. The perimeter-only rectangle has infinitely many positive real side pairs even though the controls can list only a few. The negative rectangle pair passes its numerical equation but fails the length domain; the noninteger ticket pair passes both equations but fails the count domain. Use the familiar algebraic method and the stated domain to establish the complete permitted set before interpreting what a selected candidate shows.

Worked examples

Try it yourself

1. A ribbon is 4 cm less than twice a cord's length. The cord is 9 cm long. Let r and c be their numerical lengths in centimetres. Write the relationship and find r. Would r = 2(c − 4) express the same statement?

   Show the answer and reasoning

   r = 2c − 4, giving r = 14: the ribbon is 14 cm long. 2(c − 4) is a different relationship.

   Twice the cord's length is 18 cm; subtract 4 cm to obtain 14 cm. The bracketed expression subtracts 4 first and then doubles, giving 10 cm. Reading back the whole expression catches the difference.

2. A conveyor moves at a constant 18 metres per minute for 40 seconds. How far does it move? Declare the elapsed-time unit and check the result.

   Show the answer and reasoning

   12 metres.

   Let t be the elapsed time in minutes: t = 40/60 = 2/3. Then distance is 18 × 2/3 = 12 m. The alternative consistent rate is 18/60 = 3/10 m/s, giving 3/10 × 40 = 12 m. Using 18 × 40 without conversion would treat seconds as minutes.

3. Exactly 21 tickets cost €8 each for adults and €5 each for children, with no other amounts. Can total receipts be exactly €136? Find the real algebraic pair, then explain whether rounding supplies a solution.

   Show the answer and reasoning

   The real pair is a = 31/3, c = 32/3; there is no permitted whole-ticket solution, and rounding cannot repair it.

   From a + c = 21 and 8a + 5c = 136, subtract five times the first equation: 3a = 31. Both real values are nonintegers. Rounding to (10, 11) produces €135, so it fails the exact money equation even though it keeps 21 tickets.

4. Use the ticket equations a + c = 21 and 8a + 5c = 135, with nonnegative integer counts. If at least 12 adult tickets are also required, does the complete problem have a solution?

   Show the answer and reasoning

   No. The unique equation solution is a = 10, c = 11, and it fails a ≥ 12.

   Subtract five times the count equation from the money equation to get 3a = 30, hence (10, 11). It passes the original totals and count domain, but not the extra minimum. The unique real equation pair leaves no alternative that could satisfy all conditions.

5. A rectangle has area 24 m² and length 2 m more than width. Find all real roots of the numerical area equation, then give the permitted sides.

   Show the answer and reasoning

   The real roots for width are 4 and −6. The permitted width is 4 m and length is 6 m.

   With width w m, w(w + 2) = 24 becomes (w − 4)(w + 6) = 0. The negative root gives negative sides and is rejected by the domain. The permitted pair checks both 4 × 6 = 24 square metres and 6 − 4 = 2 metres.

6. One machine finishes a freely divisible job in 4 hours and another in 12 hours. Under constant independent rates, simultaneous starts and no pauses or shared bottleneck, write the rational equation for their time together, solve it and check the original.

   Show the answer and reasoning

   For t > 0 in hours, 1/4 + 1/12 = 1/t; the completion time is exactly 3 hours.

   The left side is 1/3 job per hour, so 1/t = 1/3 and t = 3. The original denominator is nonzero. In 3 hours the machines finish 3/4 and 1/4 of the job, totalling one. Adding the individual completion times would not model their simultaneous work.

7. Positive real side lengths x m and y m satisfy x + y = 12. Does adding 2x + 2y = 24 determine one pair? What if the added condition is instead x + y = 13?

   Show the answer and reasoning

   The first addition still permits infinitely many pairs: 0 < x < 12, y = 12 − x. The second addition is contradictory and permits no pair.

   The first added equation repeats the original relationship after multiplication by 2; pairs (5, 7) and (4, 8) both pass. Subtracting x + y = 12 from x + y = 13 instead gives 0 = 1, so those two conditions cannot hold together.

What to carry forward

• Read the complete question; distinguish given data, unknowns and every exact or inequality constraint.
• Declare each variable's meaning, unit and domain before writing equations.
• Translate whole relationships and check the order and scope of operations by reading them back into words.
• Use consistent units and familiar algebra, preserving original denominator and square-root restrictions.
• Check every candidate in all original equations and all context constraints; justify that the full permitted set was considered.
• Keep exact values through the checks: rounding cannot make a fractional count solve an exact whole-object problem.
• State a meaningful answer with units and assumptions, including when data are insufficient or no solution is feasible.