Linear equations in one variable — solving, transposition method: Visual walkthrough

Step 1 — What an equation actually is: a level scale

WHAT. We start with the raw object. An equation is two expressions joined by an equals sign, like

$three copies of the unknown 3 x + a plain number 5 = a plain number 14$

Here $x$ is the variable — a box we do not yet know the contents of. The numbers $3$ , $5$ , $14$ are constants — fixed, known. The little $3$ stuck to $x$ is called the coefficient: it means "three copies of the box."

WHY start here. Before we move anything, we must agree what the " $=$ " promises. It promises the left pile and the right pile weigh exactly the same. That single promise is the engine of everything below.

PICTURE. A physical scale, level, because the two sides are equal.

Step 2 — The goal, drawn: get the box alone

WHAT. "Solving" means reaching the shape

$x = (a plain number) .$

That is the only shape where we can read off the answer. Everything we do is aimed at this target: strip away everything sitting next to $x$ until it stands alone on one pan.

WHY. Look at $3 x + 5 = 14$ . The box $x$ is buried under two things: it is multiplied by 3, and then 5 is added. To dig it out we must peel these away — but in the right order, and without tipping the scale.

PICTURE. The left pan holds three boxes plus five unit-weights; the right pan holds fourteen unit-weights. We want to end with one box alone on the left.

Step 3 — Peel off the added number (transpose the constant)

WHAT. Remove the $+ 5$ . We subtract $5$ from both pans:

$3 x + 5 - 5 = 14 - 5 ⟹ 3 x = 9.$

Term by term:

$3 x$ — the three boxes, untouched.
$- 5$ on the left cancels the $+ 5$ (five weights lifted off the left pan).
$- 5$ on the right must happen too, giving $14 - 5 = 9$ (five weights lifted off the right pan).

WHY this operation and not another. $5$ is attached to $x$ by addition. The one operation that undoes addition is subtraction — its inverse. We subtract exactly $5$ so the left pan's spare weights vanish. We could subtract $3$ or $7$ , but only $5$ makes the clutter disappear.

THE SHORTCUT (transposition). Lifting $5$ off the left and adding nothing new to the right looks, on paper, like the $+ 5$ jumped across the equals sign and became $- 5$ :

$3 x + 5 = 14 ⟹ 3 x = 14 - 5.$

That "jump-and-flip" is not a new rule — it is a photograph of "subtract 5 from both sides."

PICTURE. Five weights lifted from each pan at once; the scale stays level.

Step 4 — Peel off the coefficient (transpose the multiplier)

WHAT. We now have $3 x = 9$ : three boxes weigh nine. To find one box, split both pans into 3 equal shares:

$\frac{3 x}{3} = \frac{9}{3} ⟹ x = 3.$

Term by term:

$\frac{3 x}{3}$ — three boxes shared into three groups leaves one box per group.
$\frac{9}{3}$ — nine weights shared into three groups leaves three per group.

WHY division, and why now. The $3$ is attached to $x$ by multiplication; its inverse is division. We divide by exactly $3$ (the coefficient) so one box is left. Notice the order: we removed the added $5$ first, then divided. That is the reverse of how the expression was built ("multiply by 3, then add 5" → undo as "subtract 5, then divide by 3").

THE SHORTCUT. The $\times 3$ on the left crosses over as $\div 3$ : $3 x = 9 \Rightarrow x = \frac{9}{3}$ .

PICTURE. Each pan cut into three equal groups; compare one group to one group.

Step 5 — Check by walking back (verification)

WHAT. Put the found value $x = 3$ back into the original equation:

$3 (3) + 5 = 9 + 5 = 14. ✓$

WHY. Solving flowed forwards; checking flows backwards. If the number we dug out truly balances the scale, the two sides land on the same value. If they don't, a sign flipped somewhere.

PICTURE. The box's lid opens, three weights sit inside; three boxes (9) plus five (14) exactly matches the right pan.

Step 6 — Edge case: a negative coefficient

WHAT. What if peeling leaves a minus in front of $x$ ? Solve $12 - 2 x = 4$ .

Transpose the $12$ (it is added, so it crosses as $- 12$ ): $- 2 x = 4 - 12 = - 8.$ Now the coefficient is $- 2$ . Divide both sides by $- 2$ : $x = \frac{- 8}{- 2} = 4.$

WHY the answer is positive. A negative divided by a negative is positive — the two minus signs cancel like two "opposite" tilts undoing each other. On the scale, " $- 2 x$ " means the boxes sit on the lifting side; dividing by $- 2$ both isolates one box and flips which side it's counted on.

PICTURE. The lever with boxes on the opposite arm; dividing by the negative flips it upright.

Step 7 — Degenerate cases: when the scale refuses to answer

WHAT. Two strange things can happen at the very end.

Take the general two-sided form $a x + b = c x + d$ . Transposing gives $(a - c) x = d - b .$

If $a \neq = c$ : one clean answer, $x = \frac{d - b}{a - c}$ .
If $a = c$ and $b = d$ : you get $0 = 0$ — always true, so every number works (an identity).
If $a = c$ and $b \neq = d$ : you get something like $0 = 5$ — never true, so no number works (a contradiction).

WHY. When $a = c$ the boxes cancel entirely; the $x$ disappears. With no box left, the scale is no longer asking about $x$ — it is simply asserting whether two numbers are equal. If they already were ( $b = d$ ), any box fits; if they never were ( $b \neq = d$ ), no box fits.

PICTURE. Left: boxes cancel, weights match → any box balances. Right: boxes cancel, weights differ → permanently tipped.

The one-picture summary

Here is the entire journey in a single frame: start balanced → lift the constant off both pans → share both pans into equal groups → read one box. Every arrow is "the same move on both sides."

Recall Feynman retelling (to a 12-year-old)

Picture a see-saw that's perfectly level — that's the " $=$ ". On the left sit three mystery boxes plus five marbles; on the right sit fourteen marbles. The see-saw is level, so the boxes-and-marbles on the left weigh the same as the marbles on the right.

I want to know how heavy one box is. First I clear the marbles off the left: I lift five off the left, and — to keep it level — five off the right too. Now three boxes balance nine marbles. Then I split everything into three fair piles: one box balances three marbles. Cracked it — each box weighs three!

The "transposition" trick is just a fast way to write those lifts: a $+ 5$ that jumps the equals sign lands as $- 5$ , and a $\times 3$ that jumps lands as $\div 3$ . It looks like magic teleportation, but really I'm always doing the same thing to both sides so the see-saw never tips.

Two weird endings: if the boxes cancel and the marbles already match, any box works. If the boxes cancel but the marbles don't match, no box can ever balance it. Everything else has one neat answer.

Connections

Linear equations in one variable — solving, transposition method — the parent this deep-dive illustrates
2.1.01-Variables-and-constants — the box ( $x$ ) versus the marbles (constants)
2.1.06-Forming-equations-from-word-problems — where the scale comes from
2.1.08-Linear-equations-in-two-variables — two boxes, two unknowns
2.3.01-Linear-inequalities — when the scale isn't level
2.5.02-Balancing-chemical-equations — the same balancing logic in chemistry
3.2.04-Graphical-solution-of-linear-equations — the scale as a picture of crossing lines

Practice Recall

When

+ 5

crosses the equals sign, what does it become?

- 5

— because "crossing" is shorthand for subtracting

5

from both sides.

Why do we remove the added constant before dividing by the coefficient?

Because we undo operations in reverse of how they were built — the expression did "multiply then add," so we undo "subtract then divide."

Solve

12 - 2 x = 4

and say why the answer is positive.

x = \frac{- 8}{- 2} = 4

; negative divided by negative is positive.

For

a x + b = c x + d

with

a = c

but

b \neq = d

, how many solutions?

None — the boxes cancel and you get a false statement like

0 = 5