Equations as balanced relationships

Preserve the same solutions

An equation says two expressions have the same value. Solving it means finding every allowed input that makes that statement true.

For 3x + 5 = 20, subtract five from both sides:

Then divide both sides by the nonzero number three: x = 5. Substitute in the original equation: 3(5) + 5 = 20.

The phrase “move five to the other side” can hide the reason. Nothing moves by itself: the same subtraction is applied to both sides. The transformed equation has the same solutions because the operation is reversible.

Brackets can be removed in different valid orders

Solve 3(x-2) = 15.

Divide first: divide both sides by three to obtain x-2 = 5, then add two to obtain x = 7.

Expand first: write 3x-6 = 15, add six to obtain 3x = 21, then divide by three to obtain x = 7.

Both are valid. Choose the shorter route when you can explain why it preserves equality. Check the result in the original bracket: 3(7-2) = 15.

Check your understanding 1 Show the balancing operation

Solve -2(x-3) = 10. Give a line that shows the operation on both sides, then check your answer.

Equation steps and substitution check

Show answer

Divide both sides by -2: x-3 = -5. Add three to both sides: x = -2. Check: -2(-2-3) = -2(-5) = 10. Alternatively, expand to -2x + 6 = 10 and subtract six before dividing. The final value and the validity of each transformation are different things to check.

Remove fractional coefficients without changing the equation

Solve x/3 + 1/2 = 5/2. Multiplying every term by six gives

Subtract three and divide by two: 2x = 12, so x = 6. Check: 6/3 + 1/2 = 2 + 1/2 = 5/2.

Six was chosen because every denominator divides it. Multiplying only the x/3 term would change the equation. Another valid route subtracts 1/2 first, giving x/3 = 2, then multiplies by three.

An equation need not have exactly one solution

For 2(x + 3) = 2x + 6, expansion gives the same expression on both sides. Every real x is a solution.

For 2(x + 3) = 2x + 5, subtracting 2x gives 6 = 5. No real x makes that true, so there is no solution.

Do not divide by a variable expression without considering whether it can be zero. From x(x-1) = 0, dividing by x would discard x = 0. This lesson’s balancing with constant nonzero divisors avoids that problem.

Turn a condition into an equation

A taxi journey costs a fixed 4 dollars plus 2 dollars for each kilometre. The fare is 18 dollars. Let d be the distance in kilometres. The model is 4 + 2d = 18, so d = 7.

The conclusion needs the quantity: the journey was 7 km. Check both the model and the arithmetic: 4 + 2(7) = 18, and a nonnegative distance fits the context. A valid calculation cannot repair an equation that models the wrong fare.

Check your understanding 2 The unknown appears on both sides

A rectangle has width w cm and length (w + 3) cm. Its perimeter is 30 cm. Form an equation, solve it and state both dimensions.

Your equation, solution and dimensional check

Show answer

2w + 2(w + 3) = 30, so 4w + 6 = 30, 4w = 24 and w = 6. The width is 6 cm and the length is 9 cm. Check: 2(6) + 2(9) = 30 cm. Writing w + (w + 3) = 30 would model only half the perimeter.

Before you move on

Cover the solutions and solve one bracket equation and one equation with a fractional coefficient. If a sign or fraction step is unclear, revisit that step before a fresh check. Reading or revising a worked solution is useful practice; it does not turn the first response into independent evidence.