Two Steps, Undone in Reverse

A number machine ran two steps: times 3, then plus 5. Out came 20. To find what went in, walk backwards through the machine — undo the last step first. Take away 5: that leaves 15. Now undo times 3: split 15 into 3 equal parts. The number that went in was 5.

A two-step equation records two operations done to x. In 3x + 5 = 20, x was multiplied by 3 first, then 5 was added. To solve, apply the inverse of each operation to both sides — in reverse order. Subtract 5 from both sides: 3x = 15. Then divide both sides by 3: x = 5.

The reverse order matters because the + 5 sits outside the whole product 3x. Until it is removed, dividing by 3 hits the 5 as well and produces fractions. Undo the outermost operation first, then the one wrapped inside. Finish by checking: 3(5) + 5 = 20. It holds.

Reading 3x + 5, the order of operations says: multiply first, add second. Solving reverses that reading — subtract first, divide second — because each balance move must strip away whatever stands furthest from x. The addition was applied to all of 3x, so it is the outer layer; the multiplication is the inner one.

Both routes are legal; only one is efficient. Dividing 3x + 5 = 20 by 3 gives x + 5/3 = 20/3 — true, and solvable, but through fractions. Equation solving rewards a habit: before moving anything, say what was done to x and in what order. The solution path is that list, reversed.

You put on socks, then shoes. To undo it, the shoes come off first, then the socks. What went on last comes off first.

A number was doubled, then 2 more joined, and it became 10. Take the 2 away: 8. Split into halves: 4. The hidden number was 4.

Write f(x) = 3x + 5. Solving 3x + 5 = 20 is computing f⁻¹(20). Since f is the composition of g(x) = 3x followed by h(x) = x + 5, its inverse composes the individual inverses in the opposite order: f⁻¹ = g⁻¹ ∘ h⁻¹ — subtract 5, then divide by 3.

That reversal rule, (h ∘ g)⁻¹ = g⁻¹ ∘ h⁻¹, is completely general — it holds for any invertible processes, not only linear ones. A map x ↦ ax + b is invertible exactly when a ≠ 0, which is why every equation of this shape has exactly one solution.

The maps x ↦ ax + b with a ≠ 0 form a group under composition — the one-dimensional affine group. Solving ax + b = c applies the inverse element, and the identity (στ)⁻¹ = τ⁻¹σ⁻¹ is the algebraic content of 'last applied, first undone.' The group is nonabelian: adding then scaling differs from scaling then adding, which is why order matters at all.

The same skeleton solves Ax + b = c for matrices, conjugation problems in group theory, and layered inversions everywhere. Algebra I simply meets it first, in the smallest noncommutative setting on offer.