Quick questions on Solving equations after substitution for HSC Maths Standard 2

After substituting, collect the known numbers into a single coefficient before you start undoing operations. In $A = \tfrac{1}{2}bh$ with $A = 96$ and $h = 12$, substituting gives $96 = \tfrac{1}{2} \times b \times 12$; tidy the right side to $96 = 6b$ first, and the solve is a single division. In the trapezium formula $A = \tfrac{1}{2}(a + b)h$ with the parallel sides known, work out the bracket and the half before touching the unknown. Simplifying first turns an equation that looks complicated into one of the one-step or two-step forms you already recognise, and it removes most of the places a sign or a factor can go astray.

Once the knowns are in and tidied, the equation is linear in the unknown and you solve it exactly as in the solving-linear-equations dot point: peel the operations off the unknown in reverse order, doing the same operation to both sides to keep it balanced. If the unknown has been multiplied and then had a number added, undo the addition first and the multiplication second. The hire-cost stage sequence further down shows this happening one line at a time after the substitution.

When the unknown has more than one operation on it, the solve after substituting is a two-step (or multi-step) equation, undone in reverse order. The four panels below work through a hire-cost formula end to end. A landscaping firm hires out a turf cutter for a $120 fixed fee plus $45 an hour, so a hire of $t$ hours costs $C = 45t + 120$ dollars. A customer was charged $480; the panels find how many hours $t$ they hired it for.

Every equation you solve after substitution can be verified the same way. Put your answer back into the original formula (with all the known values in place), evaluate it, and confirm it produces the result the question gave. For the hire-cost example, $t = 8$ gives $45 \times 8 + 120 = 480$, the stated charge, so the answer is right. The check costs a few seconds and catches almost every slip, whether in the substitution or in the solving, which is why it is worth doing in the exam.

In $I = Prn$ a rate of $4\%$ must go in as $0.04$, not $4$. Forgetting this scales the answer by $100$.