Algebra represents relationships so that unknown quantities can be found or compared. Before manipulating an expression, define the variable, its units and any restrictions. A correct calculation on the wrong equation will still give the wrong answer to the original problem.
Reason through the task
Translate one relationship at a time. Distinguish a fixed charge from a per-unit charge, a percentage from its reference quantity, and a total from a difference. Keep fractions exact when possible. After solving, substitute into the original relationship and check whether the result is allowed: a length must be positive, a denominator nonzero and a count appropriately integral. Checking the modeled situation can reject a root that satisfies a transformed equation but not the original.
Worked example and a tempting error
Original example: equipment hire costs twelve dollars plus three dollars per hour. A total charge of thirty dollars gives 12 + 3h = 30. Subtract twelve to get 3h = 18, then divide by three: h = 6 hours. Substitution gives 12 + 3(6) = 30. Dividing thirty by three to obtain ten hours ignores the fixed charge. Writing 3(h + 12) changes twelve dollars into twelve additional hours and mixes units.
Try a different case
Transfer: solve (x + 1)/(x − 2) = 2. The denominator requires x ≠ 2. Multiplication gives x + 1 = 2x − 4, hence x = 5. Substituting yields 6/3 = 2, so five is valid. The restriction was recorded before multiplying because multiplying by a variable expression can otherwise conceal a value the original equation never allowed.