2 Step Equations
Theory
A two-step equation needs two inverse operations. Undo them in the reverse of the order they were done, one per line, on both sides: in \(2x - 8 = 12\), add \(8\) first (\(2x = 20\)), then divide by \(2\) (\(x = 10\)). This Year 8 maths page shows how.
Order matters. Undo addition or subtraction first, then multiplication or division, because the loose number was the last thing done to the pronumeral. Do each step to both sides.
A loose number is undone first. In \(\dfrac{x}{4} + 3 = 8\), the \(+\,3\) sits outside the fraction, so subtract \(3\), then multiply by \(4\): \(x = 20\).
A fraction line groups. In \(\dfrac{x + 3}{4} = 8\), the whole of \(x + 3\) was divided by \(4\), so the \(\div 4\) was done last. Multiply by \(4\) first, then subtract \(3\): \(x = 29\).
Two multiply-or-divide steps. \(-\dfrac{2x}{3} = 10\) multiplies \(x\) by \(-2\), then divides by \(3\). Multiply both sides by \(3\) to get \(-2x = 30\), then divide both sides by \(-2\): \(x = -15\).
Keep the sign with the pronumeral term. In \(11 - 4x = 20\) the term is \(-4x\), so after subtracting \(11\) you divide by \(-4\).
Pronumerals on both sides. Take the smaller pronumeral term away from both sides, so one side has none left. If that term is subtracted, add it instead: \(20 - 2x = 3x\) becomes \(20 = 5x\), so \(x = 4\).
Not every answer is a whole number. A decimal or a fraction is a perfectly good solution.
Testing a value. Substitute it into the left-hand side (LHS) and the right-hand side (RHS) separately. If they are equal it is the solution; if not, it is not.
Two steps, in the reverse order:
| Equation | Steps (both sides) | Solution |
|---|---|---|
| \(6a - 11 = 7\) | \(+\,11\), then \(\div 6\) | \(a = 3\) |
| \(\dfrac{m}{7} + 4 = 1\) | \(-\,4\), then \(\times 7\) | \(m = -21\) |
| \(\dfrac{k + 8}{3} = 5\) | \(\times 3\), then \(-\,8\) | \(k = 7\) |
| \(5t = 2t + 21\) | \(-\,2t\), then \(\div 3\) | \(t = 7\) |
How to solve a two-step equation
- See what was done to the pronumeral, and in what order.
- Undo the last operation first, on both sides (usually the loose \(+\) or \(-\)).
- Undo the other operation, on both sides.
- Check: substitute and show LHS = RHS.
When the pronumeral is on both sides
- Take away the smaller pronumeral term from both sides (or add it if it is subtracted).
- Solve what is left as a one-step or two-step equation, and write the pronumeral on the left.
(i) \(2x - 8 = 12\)
(ii) \(\dfrac{x}{3} + 5 = -7\)
| \(2x - 8 + 8\) | \(=\) | \(12 + 8\) |
| \(2x\) | \(=\) | \(20\) |
| \(\dfrac{2x}{2}\) | \(=\) | \(\dfrac{20}{2}\) |
| \(x\) | \(=\) | \(10\) |
| \(\dfrac{x}{3} + 5 - 5\) | \(=\) | \(-7 - 5\) |
| \(\dfrac{x}{3}\) | \(=\) | \(-12\) |
| \(3 \times \dfrac{x}{3}\) | \(=\) | \(-12 \times 3\) |
| \(x\) | \(=\) | \(-36\) |
(i) \(x = 10\)
(ii) \(x = -36\). Check: \(\dfrac{-36}{3} + 5 = -12 + 5 = -7\).
(i) \(4c + 19 = 3\)
(ii) \(\dfrac{h}{5} - 7 = -2\)
| \(4c\) | \(=\) | \(3 - 19 = -16\) |
| \(c\) | \(=\) | \(\dfrac{-16}{4} = -4\) |
| \(\dfrac{h}{5}\) | \(=\) | \(-2 + 7 = 5\) |
| \(h\) | \(=\) | \(5 \times 5 = 25\) |
(i) \(c = -4\). Check: \(4 \times (-4) + 19 = 3\).
(ii) \(h = 25\). Check: \(\dfrac{25}{5} - 7 = -2\).
The pronumeral term is \(-4x\), with \(11\) added. Subtract \(11\) from both sides, then divide both sides by \(-4\).
| \(11 - 4x - 11\) | \(=\) | \(20 - 11\) |
| \(-4x\) | \(=\) | \(9\) |
| \(\dfrac{-4x}{-4}\) | \(=\) | \(\dfrac{9}{-4}\) |
| \(x\) | \(=\) | \(-2.25\) |
\(x = -2.25\). Check: \(11 - 4 \times (-2.25) = 11 + 9 = 20\).
(i) \(\dfrac{w - 9}{4} = -3\)
(ii) \(-\dfrac{2x}{3} = 10\)
(i) The fraction line groups \(w - 9\), so multiply by \(4\) first, then add \(9\).
(ii) Multiply both sides by \(3\), then divide both sides by \(-2\).
| \(w - 9\) | \(=\) | \(-3 \times 4 = -12\) |
| \(w\) | \(=\) | \(-12 + 9 = -3\) |
| \(-2x\) | \(=\) | \(10 \times 3 = 30\) |
| \(x\) | \(=\) | \(\dfrac{30}{-2} = -15\) |
(i) \(w = -3\). Check: \(\dfrac{-3 - 9}{4} = \dfrac{-12}{4} = -3\).
(ii) \(x = -15\). Check: \(-\dfrac{2 \times (-15)}{3} = \dfrac{30}{3} = 10\).
Let \(n\) be the number of hours. Set the two costs equal, then take the smaller term, \(11n\), from both sides.
| \(11n + 24\) | \(=\) | \(15n\) |
| \(24\) | \(=\) | \(4n\) |
| \(n\) | \(=\) | \(6\) |
The cost is the same after \(6\) hours: \(11 \times 6 + 24 = 90\) and \(15 \times 6 = 90\), so \(\$90\) either way.
(ii) Which of \(x = -3\), \(x = -1\) and \(x = 2\) is the solution of \(5 - 2x = 3x + 20\)?
| \(60 - 4.5w\) | \(=\) | \(28.5\) |
| \(4.5w\) | \(=\) | \(60 - 28.5 = 31.5\) |
| \(w\) | \(=\) | \(\dfrac{31.5}{4.5} = 7\) |
(i) \(60 - 4.5w = 28.5\), so \(w = 7\) weeks.
(ii) \(x = -3\): LHS \(= 5 + 6 = 11\), RHS \(= -9 + 20 = 11\), equal. \(x = -1\) gives \(7\) and \(17\); \(x = 2\) gives \(1\) and \(26\). So \(x = -3\) is the solution.
Common pitfalls
Frequently asked questions
What is a two-step equation?
An equation that needs two inverse operations to solve, such as \(3x + 2 = 14\).
Which step do I do first?
Undo the operation that was done last, usually the added or subtracted number, then the multiplying or dividing.
How do I solve x/4 + 3 = 8?
Subtract \(3\) from both sides to get \(\dfrac{x}{4} = 5\), then multiply both sides by \(4\): \(x = 20\).
What if the pronumeral is on both sides?
Take the smaller pronumeral term away from both sides first, then solve what is left.
How do I check my answer?
Substitute it into the original equation and work out the LHS and the RHS separately; they must be equal.