Word problems

A word problem hides a pair of linear equations inside a paragraph of English. Someone's age, the cost of pens and notebooks, the digits of a two-digit number, two unknown speeds — whatever the story, the task is the same: name the two unknowns, translate each sentence into an equation, and solve the resulting simultaneous pair by substitution, elimination, or cross-multiplication. Nothing new is being tested algebraically here — every solving technique was already covered when you learned simultaneous equations on their own. What is new, and what actually separates a confident student from a stuck one, is the translation step: reading "5 less than a number" and writing $x-5$ , not $5-x$ ; reading "the sum" and reaching for a plus sign, not a times sign. This article tightens that translation skill, generalises the elimination method behind it, and works two linked examples end to end.

The core insight

Every word problem with two unknowns reduces to naming two variables, translating each sentence into one linear equation, and solving the resulting pair — so the real skill is faithful translation, not new algebra.

The core idea

Name the unknowns precisely. Vague variables cause vague equations. Instead of "let $x$ be the age," write "let the father's present age be $x$ years and the son's present age be $y$ years" — a sentence, not just a symbol.

Translate keyword by keyword. Certain English phrases map to fixed algebraic operations, and getting the direction right matters more than getting the vocabulary right:

"sum," "total," "together" $\to$ addition ( $x+y$ )
"difference," "exceeds by" $\to$ subtraction, keeping the larger quantity first
"twice," "thrice," " $k$ times as much" $\to$ multiplication by that constant
" $n$ less than $x$ " $\to \; x-n$ (not $n-x$ — this reversal is the single most common slip)
" $n$ more than $x$ " $\to \; x+n$

Count your equations before you solve. Two unknowns need exactly two independent equations. A problem sentence that only restates information you already used gives you nothing new — check each equation actually adds a fact.

Solve, then interpret. Once you have

a_1x+b_1y=c_1, \qquad a_2x+b_2y=c_2,

where: $x,y$ are the unknown quantities, $a_1,b_1,a_2,b_2$ are the numerical coefficients that come from the translation, and $c_1,c_2$ are the constants on the right — solve by substitution, elimination, or cross-multiplication (all three were built for exactly this shape of pair). The final step is easy to forget: read the numbers back into the story and reject any solution that makes no sense there — a negative number of notebooks, a fractional number of people, an age greater than 100. Four families recur constantly at this level: two numbers or ages related by a sum and a difference (or a condition set in the past/future); the cost of two kinds of items bought in two different combinations; the digits of a two-digit number, where the number itself equals $10\times(\text{tens digit})+(\text{units digit})$ , and a condition links the digits or the number to its digit-reversal; and simple speed–distance–time setups with two unknown speeds or times. The translation process below is identical across all four — only the sentences change.

Figure

A four step flow from a word problem through naming variables and translating sentences to solving and interpreting the answer

Derivation ASolving a general two-unknown word-problem pair

The goal is to show that once a word problem has been translated into two linear equations $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$ , the elimination method always produces a single formula for each unknown, which is exactly why the same three algebraic techniques — substitution, elimination, cross-multiplication — solve every word problem in this chapter regardless of the story attached to it.

Step 1. Write the two translated equations side by side:

a_1x+b_1y=c_1 \qquad (i)

a_2x+b_2y=c_2 \qquad (ii)

where: $a_1,b_1,a_2,b_2,c_1,c_2$ are the known numbers that came from translation, and $x,y$ are the unknowns the problem asks for.

Step 2. Make the $y$ -coefficients match so they can be eliminated. Multiply (i) by $b_2$ and (ii) by $b_1$ :

a_1b_2x+b_1b_2y=c_1b_2 \qquad (iii)

a_2b_1x+b_1b_2y=c_2b_1 \qquad (iv)

Step 3. Subtract (iv) from (iii); the $y$ terms cancel exactly because both now carry the coefficient $b_1b_2$ :

(a_1b_2-a_2b_1)x=c_1b_2-c_2b_1 \implies x=\frac{c_1b_2-c_2b_1}{a_1b_2-a_2b_1}.

Step 4. Repeat the same elimination on $x$ instead of $y$ (multiply (i) by $a_2$ , (ii) by $a_1$ , subtract) to get the matching formula for $y$ :

y=\frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1}.

This pair of formulas is the cross-multiplication method in disguise — you rarely need to memorise it, because repeating Steps 2–3 by hand on the actual numbers each time is just as fast and far less error-prone, but it proves that any correctly translated pair of word-problem equations has exactly one solution whenever $a_1b_2-a_2b_1\neq0$ .

Worked example

The cost of 3 pens and 2 notebooks is ₹58, while the cost of 5 pens and 4 notebooks is ₹100. Find the cost of one pen and one notebook.

Step 1. Name the unknowns. Let one pen cost ₹ $x$ and one notebook cost ₹ $y$ .

Step 2. Translate both sentences into equations.

3x+2y=58 \qquad (i)

5x+4y=100 \qquad (ii)

Step 3. Eliminate $y$ . Multiply (i) by $2$ so both equations carry $4y$ : $6x+4y=116$ . Subtracting (ii) from this, $(6x+4y)-(5x+4y)=116-100$ , gives $x=16$ .

Step 4. Substitute back into (i): $3(16)+2y=58 \implies 48+2y=58 \implies y=5$ .

Step 5. Check in the equation not used for substitution: $5(16)+4(5)=80+20=100$ ✓, matching equation (ii) exactly.

One pen costs ₹16 and one notebook costs ₹5.

Competitive stretch

A related digit problem uses the same machinery with a different translation rule. Suppose a two-digit number has digit sum $9$ , and reversing its digits gives a number $27$ greater than the original. Let the tens digit be $x$ and the units digit be $y$ , so the number is $10x+y$ and the reversed number is $10y+x$ . The two sentences give $x+y=9$ and $(10y+x)-(10x+y)=27$ , which simplifies to $9y-9x=27$ , i.e. $y-x=3$ . Adding this to $x+y=9$ : $2y=12 \implies y=6$ , then $x=3$ . The number is $\mathbf{36}$ — check: digit sum $3+6=9$ ✓, and reversed $63-36=27$ ✓.

Common traps

Writing " $n$ less than $x$ " as $n-x$ instead of the correct $x-n$ — the single most common translation error in this chapter.
Using one letter for two genuinely different unknowns, or defining a variable vaguely ("let $x$ be the age") without saying whose age or which quantity.
Setting up only one equation when the problem gives two independent facts — always count that you have two equations for two unknowns before attempting to solve.
Forgetting the interpret-and-check step: not verifying the solved values against both original sentences, and not rejecting an algebraically valid but real-world-impossible answer (negative age, fractional number of items).

Ask the tutor: If a tricky sentence like "six years ago the father's age was four times the son's" is tripping you up, ask your AI tutor to split it into "six years ago" and "four times" separately and translate each piece before combining them.

How do I know which technique to use once I have my two equations — substitution, elimination, or cross-multiplication?

All three always work, so pick whichever is fastest for the numbers in front of you. If one equation already isolates a variable (like $x=3y-12$ ), substitution is quickest. If the coefficients of one variable match or are easy multiples of each other, elimination is fastest, as in the pens-and-notebooks example. Cross-multiplication is a fixed formula that works every time but is slower to set up by hand, so save it for when the coefficients are awkward and elimination would need messy multipliers.

My two equations gave me a negative or fractional answer — did I make an algebra mistake?

Not necessarily an algebra mistake, but it is a signal to stop and check. First recheck your translation of both sentences — a sign error there (like $5-x$ instead of $x-5$ ) is the usual cause. If the translation and algebra are both correct but the answer is still negative or fractional in a context that demands a positive whole number (people, items, years), the problem itself may be inconsistent, or you may have mixed up which quantity is which in the story. Always substitute your answer back into the original sentences, in words, before trusting it.

Why do digit problems use 10x + y for a two-digit number instead of just xy?

Because $x$ and $y$ here are digits — single numbers from 0 to 9 — not the two-digit number itself, and " $xy$ " written together would just mean $x$ multiplied by $y$ , which is not what a two-digit number means. If the tens digit is $x$ and the units digit is $y$ , the tens digit contributes ten times its face value to the number's actual worth, so the number equals $10x+y$ . This is the same place-value idea you use when you write $47$ as $4\times10+7$ ; writing it algebraically just replaces the specific digits with variables.

Word problems

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads