Solving Linear Equations

Activities to try

1. 1

   Balance-scale solver: keep both pans level

   Why it works: The only way to keep the scale level while stripping it down to a single x is to mirror every move on both sides — that mirroring IS the solving method, and a negative result is a legitimate, visible outcome.

   1. Represent an equation as two pans: unknown-boxes (x) and unit-weights, using 'anti-weights' (zero-pairs) for any negative amount. Set up 8x − 6 = 2x with the −6 shown as six anti-weights on that pan.
   2. Choose an action (add or remove the same number of units, using zero-pairs where a pan has none to remove; remove an x-box; halve everything) and apply it to BOTH pans at once; watch the scale stay level.
   3. Try a one-sided action on purpose and observe that the scale tips — that move is not allowed because it breaks the equality.
   4. Solve 8x − 6 = 2x: remove 2 x-boxes from each pan to reach 6x − 6 = 0, then add 6 units to each pan (cancelling the six anti-weights) to reach 6x = 6, then divide both pans into 6 equal groups (÷6) to reach x = 1.
   5. Use an anti-x-box for −x, plus unit zero-pairs, to model 5 − x = 6: remove 5 units from both pans to reach −x = 1, then swap every item on both pans for its opposite (multiply both sides by −1) to reach x = −1, so the negative solution is something you build and see rather than distrust.
2. 2

   One true equality per line: fix the run-on equals sign

   Why it works: '=' means the two sides have equal value, not 'and then'. Every line of correct working must be a standalone true statement.

   1. Read aloud, left to right, the chain 5 − x = 6 = x = −1. Decide whether each '=' states a genuine equality (does 6 really equal x?).
   2. Rewrite the working as a vertical chain where every single line is itself true: 5 − x = 6, then −x = 1, then x = −1.
   3. Do the same for 3x − 5 = 2x: rewrite the run-on 3x − 5 = 2x = x = 5 as separate true lines, subtracting 2x from both sides first (3x − 5 = 2x → x − 5 = 0 → x = 5), and check by substitution.
   4. Spot the invalid line: given a worked solution containing a run-on chain, identify which '=' is false and correct it.
3. 3

   Solve, then check by substitution; classify how many solutions

   Why it works: A solved equation must survive a substitution check in the original; and not every equation has exactly one solution — some have none and some have infinitely many.

   1. Solve a set of equations by inverse operations, writing one true equality per line: start with 5 − x = 6 and 8x − 6 = 2x, then extend to equations with brackets and fractional/negative coefficients.
   2. For each solution, substitute it back into the ORIGINAL equation and confirm both sides are equal.
   3. Solve 2(x + 3) = 2x + 6 and decide what happens: after expanding the bracket, both sides are identical (2x + 6 = 2x + 6), so every value of x works — infinitely many solutions.
   4. Construct or examine an equation that simplifies to a false statement (e.g. a contradiction) and recognise it has no solution.
   5. Sort equations into 'one solution', 'no solution', 'infinitely many solutions'.