Word problems using linear equations: Visual walkthrough

Step 1 — The pile of notes, with nothing named yet

WHAT. Picture a piggy bank cracked open onto a table. Two kinds of paper notes fall out: some worth ₹5 each, some worth ₹10 each. We are told two facts and nothing else:

the count of notes altogether is 23,
the total money is ₹155.

WHY start here. Before any letter or equation, you must be able to see the thing the problem is about. A "word problem" is just a real situation someone wrote in sentences. If we can draw it, we can trust every symbol we invent later — each symbol will point at something in this picture.

PICTURE. Two groups of rectangles: a blue group (the ₹5 notes) and a yellow group (the ₹10 notes). We do not yet know how many are in each group — that is exactly the mystery.

Figure — Word problems using linear equations

Step 2 — Name the mystery: let $x$ be the blue count

WHAT. Give the size of the blue group a name. Write $x$ = number of ₹5 notes.

WHY a letter. A letter is a placeholder for a number we haven't found yet. It behaves like a real number in every calculation — we just don't know which number. Naming it lets us do arithmetic with the mystery instead of guessing values one at a time.

Why call it $x$ and not the yellow count? Either works. We pick the blue group; the choice is free, and we'll see the equation comes out just as cleanly.

PICTURE. The blue group now wears a label $x$ . It's still the same unknown pile — we've just tagged it.

Step 3 — Build the yellow count out of $x$

WHAT. The two groups together are 23 notes. So the yellow (₹10) group is whatever is left over after removing the blue ones: $yellow count = 23 - x .$

WHY subtract. "Together they make 23" is a total, and a total is a sum: (blue) + (yellow) = 23. To get yellow alone we peel the blue off the total — that peeling-off is subtraction, the inverse operation of adding. This is the whole point of Step 2 of the five-step method: express every other quantity using the one name you chose.

PICTURE. A bar of length 23 split into a blue chunk of length $x$ and a yellow chunk that fills the rest — clearly labelled $23 - x$ .

Step 4 — Turn counts into money

WHAT. Money, not counts, is our second fact. Each blue note is worth ₹5, so the blue pile is worth $5 \times x = 5 x rupees .$ Each yellow note is worth ₹10, so the yellow pile is worth $10 \times (23 - x) = 10 (23 - x) rupees .$

WHY multiply. "Worth ₹5 each" is a rate: rupees per note. To turn "per note" into total rupees you multiply the rate by how many notes there are. That is what multiplication is — repeated adding of the same value. Term by term:

$₹ per blue note 5 \times how many blue x = ₹ from all blue notes 5 x$

$₹ per yellow note 10 \times how many yellow (23 - x) = ₹ from all yellow notes 10 (23 - x)$

PICTURE. Each note rectangle now carries its value; the blue stack labelled $5 x$ , the yellow stack labelled $10 (23 - x)$ . Height = money.

Step 5 — Set the money equation

WHAT. The two money-piles combined equal the known total, ₹155: $5 x + 10 (23 - x) = 155.$

WHY this is the equation. Value is extensive — separate piles of money simply add up, no interaction between them. So (money from blue) + (money from yellow) = (total money). The word "is / total" in the problem is the English for the " $=$ " sign. Term by term:

$blue money 5 x + yellow money 10 (23 - x) = total money, given 155$

PICTURE. A balance scale: left pan holds the blue bar plus the yellow bar; right pan holds a single ₹155 weight. They balance — that balance is the equals sign.

Step 6 — Unwrap the brackets

WHAT. Multiply the $10$ into the bracket: $5 x + 230 - 10 x = 155.$

WHY. $10 (23 - x)$ means "ten copies of $(23 - x)$ ", which is ten copies of $23$ minus ten copies of $x$ . Spreading a multiplier across a sum is the distributive law — the backbone of working with Algebraic Expressions: $10 \times 23 = 230, 10 \times (- x) = - 10 x .$

PICTURE. The yellow bar $10 (23 - x)$ literally split into a fixed green slab of height $230$ and a removed red slab of height $10 x$ — showing the " $+ 230 - 10 x$ ".

Step 7 — Collect and isolate $x$

WHAT. Combine the two $x$ -terms, then peel $x$ free: $5 x - 10 x + 230 = 155 \Rightarrow - 5 x + 230 = 155.$ Subtract $230$ from both sides: $- 5 x = 155 - 230 = - 75.$ Divide both sides by $- 5$ : $x = \frac{- 75}{- 5} = 15.$

WHY these moves. An equation is a balance. Whatever you do to one pan you must do to the other, or it tips. We use inverse operations: adding $230$ was in the way, so we subtract $230$ ; $x$ was multiplied by $- 5$ , so we divide by $- 5$ . Each move undoes one wrapping around $x$ until $x$ stands alone. Note $\frac{- 75}{- 5} = + 15$ : a negative divided by a negative is positive.

PICTURE. The balance scale, three frames: (1) both pans; (2) remove a $230$ block from each pan — still level; (3) shrink each pan by a factor of $5$ — landing on $x = 15$ .

Step 8 — The edge cases: does the answer even make sense?

WHAT. Two sanity checks the algebra alone won't give you:

Counting check. $x = 15$ is a whole number and $0 \leq 15 \leq 23$ . Good — you cannot have half a note or a negative pile.
Money check. $5 (15) + 10 (8) = 75 + 80 = 155$ ✓, and $15 + 8 = 23$ ✓.

WHY worry. Algebra will happily hand you $x = 15.7$ or $x = - 4$ if the story numbers were impossible. Because $x$ is a count (Step 2), only whole numbers in $[0, 23]$ are physically real. This is why Step 5 of the method — Verify and Interpret — exists.

Degenerate boundaries. What if the total money were exactly $23 \times 5 = 115$ ? Then $x = 23$ : all blue, zero yellow — a valid but extreme answer. If it were $23 \times 10 = 230$ ? Then $x = 0$ : all yellow. Any total below 115 or above 230 has no whole-note solution — the story would be impossible, and a fractional or out-of-range $x$ is your warning flag.

PICTURE. A number line for $x$ from $0$ to $23$ . Green window = valid whole-number range; our $x = 15$ sits safely inside; a ghost point at $x = - 4$ shown crossed out (impossible).

The one-picture summary

Everything above, compressed: the pile → the name $x$ → the split $x$ and $23 - x$ → money bars → the balanced equation → the peeled-away solution → the reality check.

Recall Feynman retelling — say it like a story

I had a pile of ₹5 and ₹10 notes: 23 notes worth ₹155, but I didn't know how many of each. So I pointed at the ₹5 group and said, "you're $x$ ." The rest of the 23 must be the ₹10 group, so that's $23 - x$ . Now money: the ₹5 group is worth $5 x$ , the ₹10 group is worth $10 (23 - x)$ , and money just adds up, so together they equal the ₹155 I know: $5 x + 10 (23 - x) = 155$ . I opened the bracket ( $10$ times $23$ is $230$ , $10$ times $x$ is $10 x$ ), tidied the $x$ 's to get $- 5 x + 230 = 155$ , undid the $+ 230$ by subtracting it from both sides, then undid the $\times (- 5)$ by dividing — and out popped $x = 15$ . That means 15 blue notes and 8 yellow notes. I checked: 15 and 8 make 23, and ₹75 plus ₹80 make ₹155. Both whole numbers, both sensible — done. The whole trick was: name the mystery, build everything else out of that name, balance what you know, then undo the wrapping until the mystery stands alone.

Recall Quick self-test

Why can't we use two letters $x$ and $y$ for the two note counts here? ::: With two unknowns we'd need two equations to pin them; writing yellow as $23 - x$ keeps one unknown so one equation suffices. What real fact does the " $=$ " in $5 x + 10 (23 - x) = 155$ represent? ::: The two money-piles combined are exactly the given total ₹155. After getting $x = 15.7$ in some similar problem, what should you conclude? ::: Since notes are whole, a fractional count means the story's numbers are impossible — no valid solution. What does distributing the $10$ in $10 (23 - x)$ give, term by term? ::: $230 - 10 x$ .

See also the general recipe on the parent note, and related setups in Simultaneous Linear Equations, Ratio and Proportion and Applications of Algebra.