Word problems using linear equations: Worked examples

You have met the five-step method and three clean examples where the answer came out positive and whole. But real problems are messier. What if the equation gives a negative number? A fraction? A zero? What if it gives a number that is correct algebraically but nonsense in the real world? What if the equation collapses to $0 = 0$ or $0 = 5$ ?

This page walks through every case class a linear word problem can throw at you, one fully worked example per case, so you never meet a surprise in an exam.

The scenario matrix

Every cell below is a distinct thing that can happen. Each worked example is tagged with the cell it lands in.

Cell	What happens	Danger	Example
A. Positive whole answer	$x$ comes out clean and sensible	none — the easy case	Ex 1
B. Fractional answer that is fine	$x$ is a fraction, and fractions are allowed (money, time)	rounding wrongly	Ex 2
C. Fractional answer that is nonsense	$x$ is a fraction, but must be whole (people, notes)	accepting an impossible answer	Ex 3
D. Negative answer that is valid	$x < 0$ but it means "into the past / below zero"	wrongly rejecting it	Ex 4
E. Negative answer that is impossible	$x < 0$ but the story forbids it	wrongly accepting it	Ex 5
F. Zero is the answer	$x = 0$ is a genuine, valid solution	thinking "zero means I failed"	Ex 6
G. Two unknowns, one relation each	need [[Simultaneous Linear Equations	two equations]] or a clever substitution	mixing up which is which
H. Degenerate: no solution ( $0 = k$ )	the story contradicts itself	thinking you did the algebra wrong	Ex 8
I. Degenerate: infinite solutions ( $0 = 0$ )	the story gives no real constraint	thinking there is a unique answer	Ex 9
**J. Exam twist: hidden [[Ratio and Proportion	ratio]] + limiting value**	translation is buried; check the boundary	translating the ratio backwards

The tools we lean on throughout are writing quantities as expressions in $x$ and inverse operations to isolate $x$. If any step feels fast, that is the note to revisit. To keep this visual-first, every example carries a small picture — a number line, a bar model, or a graph — so you can see the case, not just read it.

Worked example Example 1 — Cell A: clean positive whole number

Problem: The sum of three consecutive integers is $72$ . Find them.

Forecast: Guess: they should cluster around $72 \div 3 = 24$ . Hold that thought.

Let the smallest integer be $x$ . The next two are $x + 1$ and $x + 2$ . Why this step? "Consecutive" means each is one more than the last, so $+ 1$ and $+ 2$ encode that exactly.
Set up the equation from "sum is 72": $x + (x + 1) + (x + 2) = 72$ Why this step? "Sum" → add the three expressions; "is" → $=$ .
Combine like terms: $3 x + 3 = 72$ . Why this step? $x + x + x = 3 x$ and $1 + 2 = 3$ ; collecting terms is the first move toward isolating $x$.
Subtract 3, then divide by 3: $3 x = 69$ , so $x = 23$ . Why this step? Undo the $+ 3$ (inverse of add is subtract), then undo the $\times 3$ (inverse of multiply is divide).

Answer: $23, 24, 25$ .

Verify: $23 + 24 + 25 = 72$ ✓. And $24$ (the middle) matched our forecast of "around 24". ✓

Figure (below): three blocks laid end to end — of lengths $23$ , $24$ , $25$ — whose combined length is exactly $72$ . The middle block ( $24$ , in red) is the forecast average; an arrow points to it.

Worked example Example 2 — Cell B: fraction is a perfectly good answer

Problem: Pipe A fills a tank in 4 hours, Pipe B in 6 hours. Working together, how long to fill it?

Forecast: Together must be faster than the faster pipe alone, so the answer must be less than 4 hours. Watch for that.

Let the joint time be $t$ hours. Why this step? That is exactly the quantity asked for; no other unknown is needed.
Write each rate. A does $\frac{1}{4}$ tank/hour, B does $\frac{1}{6}$ tank/hour. Why this step? Rate = "fraction of job per hour". Rates add, times do not.
One tank of work in time $t$ : $t (\frac{1}{4} + \frac{1}{6}) = 1$ Why this step? $rate \times time = work$ , and we want work $= 1$ whole tank.
Add the fractions using the LCM $LCM (4, 6) = 12$ : $\frac{1}{4} = \frac{3}{12}$ and $\frac{1}{6} = \frac{2}{12}$ , so $\frac{3}{12} + \frac{2}{12} = \frac{5}{12}$ . Thus $t \cdot \frac{5}{12} = 1$ . Why this step? Fractions can only be added once they share a denominator; the LCM gives the smallest common floor, so we rewrite both over $12$ before adding.
Divide by $\frac{5}{12}$ = multiply by $\frac{12}{5}$ : $t = \frac{12}{5} = 2.4$ hours $= 2$ h $24$ min. Why this step? To isolate $t$ we undo the multiplication by $\frac{5}{12}$ ; dividing by a fraction is the same as multiplying by its reciprocal.

Answer: $\frac{12}{5}$ hours $= 2.4$ hours.

Verify: Is $2.4 < 4$ ? Yes ✓ (matches forecast). Work: $2.4 \cdot \frac{1}{4} + 2.4 \cdot \frac{1}{6} = 0.6 + 0.4 = 1$ tank ✓. A fraction here is not an error — time is continuous.

Figure (below): a bar showing the fraction of tank filled in one hour — A's slice $\frac{1}{4}$ plus B's slice $\frac{1}{6}$ — with a red mark at their sum $\frac{5}{12}$ . Because this combined rate is bigger than either alone, the fill-time $t = \frac{12}{5}$ is smaller than 4 hours.

Worked example Example 3 — Cell C: fraction is a red flag

Problem: Tickets cost ₹40 each. A group spends exactly ₹450 on tickets, buying $x$ tickets. How many tickets?

Forecast: Tickets come in whole numbers. If the algebra gives a fraction, the problem itself is broken — a good exam trap.

Let the number of tickets be $x$ (already given). Why this step? The quantity asked for is already named; no other unknown appears.
Equation "total spent is 450": $40 x = 450$ . Why this step? value-per-ticket $\times$ count $=$ total; "is" → $=$ .
Divide by 40: $x = \frac{450}{40} = 11.25$ . Why this step? Inverse of $\times 40$ is $\div 40$ , which isolates $x$ .

Answer / interpretation: $x = 11.25$ tickets is impossible — you cannot buy a quarter ticket.

Verify: $40 \times 11.25 = 450$ ✓ algebraically, but $11.25$ fails the reality check (must be a whole number). Correct response: "No whole number of ₹40 tickets costs exactly ₹450." Step 5 of the method (interpret!) is what saves you here — the algebra alone would mislead.

Figure (below): the cost of whole ticket counts ( $9, 10, 11, 12, 13$ ) plotted as dots at $360, 400, 440, 480, 520$ . A dashed line at the target $450$ crosses the staircase in the gap between $11$ and $12$ — the red point at $x = 11.25$ shows the answer lands between two whole tickets, so no whole count works.

Worked example Example 4 — Cell D: negative answer that IS valid

Problem: A thermometer reads $12°$ C now and drops $3°$ every hour. When (relative to now) did it read $21°$ C?

Forecast: $21°$ is warmer than now, and it is cooling. So it read $21°$ in the past — expect a negative time.

Let $t$ = hours from now ( $t > 0$ future, $t < 0$ past). Why this step? We deliberately allow negative $t$ so "past" is expressible.
Temperature at time $t$ : $12 - 3 t$ . Why this step? Starts at 12, loses $3$ each hour → subtract $3 t$ .
Set equal to 21: $12 - 3 t = 21$ . Why this step? The question fixes the reading at $21°$ ; "read $21°$ " → set the temperature expression equal to $21$ .
Solve: $- 3 t = 9 \Rightarrow t = - 3$ . Why this step? Subtract 12 from both sides (undo the $+ 12$ ), then divide by $- 3$ (undo the $\times (- 3)$ ); dividing a positive by a negative gives a negative.

Answer: $t = - 3$ , i.e. 3 hours ago.

Verify: $12 - 3 (- 3) = 12 + 9 = 21$ ✓. The negative is meaningful and accepted here — the number line has a past. See the figure: the red point sits to the left of "now".

Figure (below): a horizontal time-line with "now" at $t = 0$ (12 °C). The red dot sits at $t = - 3$ , to the left of now, labelled "3 h ago, 21 °C" — showing a negative answer simply lives in the past region of the line.

Worked example Example 5 — Cell E: negative answer that is IMPOSSIBLE

Problem: Reena is 8 years old. In how many years will she be $5$ years old?

Forecast: She is already older than 5. To be 5 again you'd have to go backward in time, which the phrase "in how many years" forbids (it means the future).

Let $y$ = years into the future (the phrase pins $y \geq 0$ ). Why this step? "In how many years" means future only, so we build the constraint $y \geq 0$ into the variable from the start.
Age then: $8 + y$ . Why this step? Each future year adds 1 to her age, so after $y$ years she is $8 + y$ .
Equation: $8 + y = 5$ . Why this step? "Will be 5" fixes her future age at $5$ ; "will be" → $=$ .
Solve: $y = - 3$ . Why this step? Subtract 8 from both sides (undo the $+ 8$ ) to isolate $y$ .

Answer / interpretation: $y = - 3$ contradicts $y \geq 0$ . So it never happens in the future.

Verify: $8 + (- 3) = 5$ ✓ algebraically, but $- 3$ violates the story's constraint (future only). Compare with Ex 4: same negative number, opposite verdict — the story decides, not the sign alone.

Figure (below): the same style of time-line, but now an arrow marks the allowed region "future only, $y \geq 0$ " to the right of now. The answer $y = - 3$ is drawn as a red ✗ in the banned past region — same $- 3$ as Ex 4, but here it is forbidden.

Worked example Example 6 — Cell F: zero is a real answer

Problem: A shop gives a flat ₹50 joining fee plus ₹20 per magazine. Anaya's bill was exactly ₹50. How many magazines did she buy?

Forecast: She paid only the joining fee, so she must have bought — pause and guess — none. Expect $x = 0$ .

Let $x$ = number of magazines bought. Why this step? That is the quantity asked for, and it can honestly be $0$ .
Bill expression: $50 + 20 x$ . Why this step? Fixed fee ₹50 plus ₹20 for each of $x$ magazines.
Equation: $50 + 20 x = 50$ . Why this step? "Bill was ₹50" → set the bill expression equal to $50$ .
Solve: subtract 50 → $20 x = 0$ ; divide by 20 → $x = 0$ . Why this step? Undo the $+ 50$ then the $\times 20$ ; $0$ divided by anything nonzero is $0$ .

Answer: $x = 0$ — she bought no magazines.

Verify: $50 + 20 (0) = 50$ ✓, and $0$ is a whole number $\geq 0$ , so it is perfectly valid. Zero is a genuine solution, not a sign of failure — the bar in the figure shows the "magazine" section has zero length.

Figure (below): a bill-bar of total length $50$ : the whole bar is the black "joining fee = 50" block, and the red magazine section has zero length (marked by a red tick at the far end). The empty magazine part is the picture of $x = 0$ .

Worked example Example 7 — Cell G: two unknowns, substitute one out

Problem: A piggy bank has ₹5 and ₹10 notes, 23 notes worth ₹155 in total. How many of each?

Forecast: More small notes than large, probably, since ₹155 over 23 notes averages under ₹7 per note.

Let ₹5 notes $= x$ . Then ₹10 notes $= 23 - x$ . Why this step? Two unknowns, but the total-count relation lets us write the second in terms of the first — turning two equations into one.
Value equation: $5 x + 10 (23 - x) = 155$ . Why this step? value = rate × count for each type, summed to the total value.
Expand: $5 x + 230 - 10 x = 155 \Rightarrow - 5 x + 230 = 155$ . Why this step? Multiply the $10$ into the bracket (the distributive rule) so all terms are loose, then gather the $x$ -terms $5 x - 10 x = - 5 x$ .
Solve: $- 5 x = - 75 \Rightarrow x = 15$ . So ₹10 notes $= 23 - 15 = 8$ . Why this step? Subtract 230 from both sides (undo the $+ 230$ ), then divide by $- 5$ (undo the $\times (- 5)$ ); a negative over a negative is positive.

Answer: $15$ notes of ₹5, $8$ notes of ₹10.

Verify: count $15 + 8 = 23$ ✓; value $5 (15) + 10 (8) = 75 + 80 = 155$ ✓. Average $155/23 \approx 6.7 < 7$ , matching the forecast. ✓

Figure (below): two stacked bars. The top "count" bar splits $23$ into $15$ five-rupee notes (white) and $8$ ten-rupee notes (red). The bottom "value" bar splits $155$ into $75$ (from the fives) and $80$ (from the tens, red) — one $x$ satisfies both the count total and the value total.

Worked example Example 8 — Cell H: no solution (

$0 = k$ ) Problem: "Twice a number, increased by 6, equals twice the same number increased by 10." Find the number.

Forecast: Something feels off — the two sides move together. Watch the $x$ 's cancel.

Let the number be $x$ . Why this step? There is one unknown quantity — "a number" — so we name it $x$ .
Equation: $2 x + 6 = 2 x + 10$ . Why this step? "Twice a number increased by 6" is $2 x + 6$ ; the same on the right with $10$ ; "equals" → $=$ .
Subtract $2 x$ from both sides: $6 = 10$ . Why this step? We try to collect $x$ -terms — but they vanish entirely, leaving only constants.

Answer / interpretation: $6 = 10$ is false for every $x$ , so no number works.

Verify: The $x$ -coefficients are equal ( $2 = 2$ ) but the constants differ ( $6 \neq = 10$ ) → parallel "lines that never meet." No solution — and you did the algebra correctly; the story is self-contradictory. The figure shows the two parallel lines.

Figure (below): the two sides drawn as lines $y = 2 x + 6$ (black) and $y = 2 x + 10$ (red). They have the same slope, so they are parallel and never cross — no $x$ makes the sides equal, which is what "no solution" looks like.

Worked example Example 9 — Cell I: infinitely many solutions (

$0 = 0$ ) Problem: "A number tripled and then reduced by 4 equals four less than three times the number." Find it.

Forecast: Both sides describe the same thing in different words. Expect a tautology.

Let the number be $x$ . Why this step? One unknown quantity — "a number" — so we name it $x$ .
Translate both sides. Left "a number tripled then reduced by 4" $= 3 x - 4$ . Right "four less than three times the number" $= 3 x - 4$ . Why this step? "Tripled" → $3 x$ , "reduced by 4" and "four less than" both → $- 4$ ; they say the identical thing.
Equation: $3 x - 4 = 3 x - 4$ . Why this step? "Equals" → $=$ ; both translations are the same expression.
Subtract $3 x$ from both sides: $- 4 = - 4$ . Add 4 to both sides: $0 = 0$ . Why this step? We try to isolate $x$ — undo the $3 x$ , then undo the $- 4$ — but everything cancels, leaving a statement with no $x$ left in it.

Answer / interpretation: $0 = 0$ is true for every $x$ → every number is a solution.

Verify: Pick $x = 100$ : $3 (100) - 4 = 296$ and $3 (100) - 4 = 296$ ✓. Pick $x = - 7$ : $3 (- 7) - 4 = - 25$ and $3 (- 7) - 4 = - 25$ ✓. The two descriptions were identical, so the "equation" carried no real constraint at all.

Figure (below): both sides drawn as the same line $y = 3 x - 4$ — a thick black line with a red dashed line lying exactly on top of it. Because the two lines coincide everywhere, every $x$ is a meeting point, which is what "infinitely many solutions" looks like.

Worked example Example 10 — Cell J: exam twist — hidden ratio + limiting check

Problem: Two numbers are in the ratio $3 : 5$ . If $4$ is added to each, the new ratio becomes $2 : 3$ . Find the numbers.

Forecast: Adding the same amount to both nudges the ratio toward $1 : 1$ . Since $2 : 3$ ( $\approx 0.67$ ) is closer to 1 than $3 : 5$ ( $= 0.6$ ), the setup is consistent. Good sign.

Use one variable for the whole ratio: let the numbers be $3 k$ and $5 k$ . Why this step? A ratio $3 : 5$ means "some common multiple" $k$ scales both. One unknown, not two.
After adding 4: numbers become $3 k + 4$ and $5 k + 4$ , ratio $2 : 3$ : $\frac{3 k + 4}{5 k + 4} = \frac{2}{3}$ Why this step? "New ratio is $2 : 3$ " → set the fraction of the two new numbers equal to $\frac{2}{3}$ .
Cross-multiply (an inverse operation that clears the fractions): $3 (3 k + 4) = 2 (5 k + 4)$ Why this step? Multiplying both sides by both denominators removes the fractions, leaving a plain linear equation.
Expand: $9 k + 12 = 10 k + 8$ . Why this step? Apply the distributive rule to each bracket so the $k$ -terms and constants stand free, ready to collect.
Solve: $12 - 8 = 10 k - 9 k \Rightarrow k = 4$ . Why this step? Move constants to one side and $k$ -terms to the other (inverse operations) to isolate $k$ .
Recover the numbers: $3 k = 12$ , $5 k = 20$ . Why this step? The question asked for the numbers themselves, not $k$ ; substitute $k = 4$ back into $3 k$ and $5 k$ .

Answer: $12$ and $20$ .

Verify: original ratio $12 : 20 = 3 : 5$ ✓; after $+ 4$ : $16 : 24 = 2 : 3$ ✓. Limiting sanity check: if the added amount grew without bound, the ratio would approach $1 : 1$ — and indeed $2 : 3$ is nearer $1 : 1$ than $3 : 5$ , so a finite positive $k$ is exactly what we should get. ✓

Figure (below): two stacked bars. The top "before" bar splits into $12$ and $20$ (ratio $3 : 5$ ); the bottom "after" bar shows the same $12$ and $20$ each grown by a red $+ 4$ chunk, giving $16$ and $24$ (ratio $2 : 3$ ). The equal red add-ons pull the ratio toward $1 : 1$ .

Recall Quick self-test

(Format below: the part before the divider is the prompt; the part after is the answer to reveal.) A cooling problem gives $t = - 2$ hours. Valid or not? ::: Valid — it means "2 hours ago", exactly like Ex 4, provided the story allows the past. A shop-bill problem gives $x = 0$ items. Valid or not? ::: Valid — zero is a genuine solution (Ex 6): the customer bought none. An equation reduces to $7 = 7$ . How many solutions? ::: Infinitely many — every value of the variable works (Cell I). An equation reduces to $7 = 2$ . How many solutions? ::: None — the statement is impossible (Cell H). You buy ₹40 tickets and get $x = 11.25$ . What do you report? ::: "No whole number of tickets costs exactly that" — reject the fraction (Cell C).

See also: Linear Equations in One Variable · Applications of Algebra · Word problems using linear equations