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Algebra · Formulae & rearranging

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Substituting into formulae

Why substitution is "replace the letter, keep the structure", why negatives must go in with brackets, and why the order of operations still rules once the numbers are in.

Algebra · Formulae & rearranging

Substituting into formulae

Why substitution is "replace the letter, keep the structure", why negatives must go in with brackets, and why the order of operations still rules once the numbers are in.

Why it works

A formula is a machine: letters mark the slots where numbers go. To substitute is to replace each letter with its number — brackets round it — and then evaluate with the usual order of operations. Every substitution error is a failure of one of those two clauses.

Algebra's hidden multiplications become visible. atat means a×ta \times t and 3q23q^2 means 3×q23 \times q^2 — when a=3a = 3 and t=4t = 4, at=12at = 12 (never "34"). Writing the bracket stage — v=−6+(3)(4)v = -6 + (3)(4) — makes the invisible ×\times explicit before any arithmetic happens.

The power binds before the coefficient multiplies. 3q23q^2 with q=−5q = -5: square first, then times 3 — 3×25=753 \times 25 = 75. Working (3×−5)2=225(3 \times -5)^2 = 225 multiplies too early; the coefficient is waiting OUTSIDE the square. And q2=(−5)2=+25q^2 = (-5)^2 = +25: a negative goes into the slot with its bracket, so the square sees the whole of −5-5.

Two-term formulae: evaluate each term, then combine. With u=5u = 5, a=−2a = -2, t=6t = 6:

s=ut+12at2=(5)(6)+12(−2)(36)=30−36=−6.s = ut + \tfrac{1}{2}at^2 = (5)(6) + \tfrac{1}{2}(-2)(36) = 30 - 36 = -6.

A negative result is not an error — here it just means displacement ended up behind the start. Let the arithmetic speak.

Match each value to its own letter. In v=u+atv = u + at, uu is the starting speed and vv the finishing one — swapping them is a classic slip that no amount of correct arithmetic can rescue. Read the story, list the letters with their values first, then substitute.

When the answer is squared, finish the job. v2=u2+2asv^2 = u^2 + 2as gives v2=49v^2 = 49 — the question asked for vv, so one more step: v=7v = 7. Stopping at 49 (or forgetting the root exists) hands back the wrong quantity.