How to rearrange a formula without losing a term
A formula may give one quantity in terms of several others, while your question asks for a different quantity. Rearranging means finding an equivalent equation with the required variable on its own. Write each operation on both sides so you can see where every term goes.
Name the subject and undo the outer operations
Start by writing “make __ the subject”. Circle every occurrence of that variable. If it appears once, identify the operations around it and undo them in reverse order. Keep the two sides equal at every step; a term does not change sign simply because you move it across the page.
For v = u + at, make t the subject. Subtract u from both sides to obtain v − u = at. Then divide both sides by a: t = (v − u) / a, provided a is not zero. The brackets matter. Writing v − u / a would divide only u by a.
Try a numerical check: if u = 3, a = 2 and t = 4, the original formula gives v = 11. The rearranged formula gives (11 − 3) / 2 = 4. This is a useful way to catch a lost bracket or incorrect sign, although one successful example does not prove an algebraic identity.
Treat a bracket or fraction as a whole expression
For y = 3(x + 2), make x the subject. Divide the whole equation by 3 to get y / 3 = x + 2, then subtract 2: x = y / 3 − 2. Expanding first also works, but doing so creates an extra term to track. Choose the route you can record clearly.
For q = (m + n) / p, make m the subject. Multiply both sides by p: qp = m + n. Subtract n to obtain m = qp − n. The original fraction requires p ≠ 0, and that restriction still matters even though p is no longer a denominator in the rearranged equation.
If a denominator contains several terms, keep brackets around them when multiplying. From z = k / (x + 4), multiply by x + 4 to get z(x + 4) = k. For z ≠ 0, divide by z and subtract 4: x = k / z − 4. Also retain the original restriction x ≠ −4. Check a proposed value against both restrictions.
Collect and factor when the variable appears twice
If the required variable appears in two terms, ordinary inverse operations may leave another copy behind. Bring its terms together, then factor out the variable. You can divide by the remaining factor only when it is nonzero.
Consider P = ax + bx + c. Subtract c: P − c = ax + bx. Factor the right side: P − c = x(a + b). Divide by a + b to obtain x = (P − c) / (a + b), provided a + b ≠ 0. You cannot divide just ax by a and leave bx unchanged.
When a + b = 0, return to the equation before division. It becomes P − c = 0. If P = c, every x satisfies this equation; if P ≠ c, no x does. There is no single value of x to extract by dividing by zero.
For a practice check with a = 2, b = 3, c = 7 and P = 27, the rearranged formula gives x = (27 − 7) / (2 + 3) = 4. Substituting into the original gives 2 × 4 + 3 × 4 + 7 = 27.
Use a short check before substituting numbers
Read the final expression as instructions. In t = (v − u) / a, subtract u from v first, then divide the result by a. If your calculator entry performs a different sequence, correct the entry rather than trusting the displayed number.
Use this working template: “Required subject: __. Original equation: __. Operation on both sides: __. Equation after that operation: __. Restrictions on division: __. Final expression: __. Substitution check: __.” Repeat the operation line whenever another step is needed.
If your steps involve squaring, taking roots or cancelling a factor containing the unknown, check the possible solutions in the original equation. Those operations can introduce an extra solution or remove a case. When unsure, keep the uncancelled equation and ask which values the operation excludes.
