Chapter 2.4 — Exercise 2.3 — Variables on Both Sides

A Two-Pass Strategy

Rather than transposing term by term in whatever order they appear, it helps to work in two clean passes:

• Pass one — gather the variable terms. Transpose every variable term from the right side over to the left.
• Pass two — gather the constants. Transpose every plain number from the left side over to the right.
• Then simplify each side by combining like terms, and finally divide both sides by whatever coefficient is left in front of the variable.
• If brackets are present, expand every one of them using the distributive law before starting either pass — trying to transpose through an unexpanded bracket is where most errors begin.

Worked Examples, in Order of Difficulty

A first pass with no brackets: 7x − 5 = 2x. Move 2x to the left and −5 to the right: 5x = 5, so x = 1.

Constants on both sides too: 7p − 3 = 3p + 8. Move 3p left, −3 right: 4p = 11, giving p = 11/4 — a perfectly valid fractional answer, not a sign that something went wrong.

A single bracket to expand first: 3x + 4 = 5(x − 2). Expanding the right side first is what unlocks the rest of the process:

Brackets on both sides: 15(x − 1) + 4(x + 3) = 2(7 + x). Expand all three brackets before transposing anything:

Several brackets, more terms to track: 3(5z − 7) + 2(9z − 11) = 4(8z − 7) − 111. Expanding each bracket gives 15z − 21 + 18z − 22 = 32z − 28 − 111, and collecting like terms on each side before transposing keeps this manageable: 33z − 43 = 32z − 139, so z = −96.

A subtraction of a whole bracket: 8(x − 3) − (6 − 2x) = 2x + 4 − 5(5 − x). The minus sign directly in front of (6 − 2x) has to flip both terms inside it when the bracket is removed — it becomes −6 + 2x, not −6 − 2x. Expanding fully: 8x − 24 − 6 + 2x = 2x + 4 − 25 + 5x, which simplifies to 10x − 30 = 7x − 21, giving 3x = 9 and x = 3.

Why It Doesn't Matter Which Side You Collect On

Both passes described above chose to collect variables on the left and constants on the right, but that choice is a convenience, not a rule. Take 9y + 5 = 15y − 1: moving in the other direction, transpose 9y to the right and −1 to the left, giving 6 = 6y, so y = 1 — the same answer as collecting on the left would give. The only real consequence of the direction you choose is which side ends up positive; picking whichever direction keeps the variable's coefficient positive is a small habit that avoids an extra sign flip at the very end.

One More Worked Example

5(z + 3) = 4(2z + 1) has a bracket on both sides at once. Expanding both before touching anything else: 5z + 15 = 8z + 4. Now the two passes: transpose 8z to the left and 15 to the right: −3z = −11, so z = 11/3. Notice that even with brackets on both sides, the order never changes — expand everything first, and only then start transposing.

Where the Errors Usually Are

• A sign left unflipped during transposition. Moving −5 to the other side must produce +5 — the sign always reverses when a term crosses the equals sign.
• A bracket expanded only partially. Every single term inside a bracket must be multiplied by the factor outside it — in 5(x − 2), both the x and the −2 are multiplied by 5, not just the x.
• Transposing before combining like terms. In the busier examples with several brackets, expand everything and combine like terms on each side first — only then start moving terms across the equals sign.
• A minus sign in front of a bracket, not a number. Removing −(6 − 2x) requires flipping the sign of every term inside, giving −6 + 2x — a very easy place for a slip to happen unnoticed.
• Assuming a fractional answer must be wrong. Results like 11/4 are exactly as valid as whole-number answers; there's no rule that the variable has to come out as an integer.

Checking an Equation Once the Brackets Are Involved

Verification matters even more once brackets are in play, because there are more places for a slip to hide. Take the earlier result x = 1 for 15(x − 1) + 4(x + 3) = 2(7 + x). Substituting back into the original, unexpanded equation: 15(1 − 1) + 4(1 + 3) = 15(0) + 4(4) = 0 + 16 = 16. On the right: 2(7 + 1) = 2(8) = 16. Both sides equal 16, confirming the answer. Substituting into the original form, brackets and all, rather than a partially simplified version, is what actually catches an expansion mistake — checking against an already-expanded line just repeats whatever error might already be in it. The extra minute this takes is small compared to the time lost tracking down a wrong answer after the fact, especially once an equation has three or four brackets rather than just one, where a single dropped sign can otherwise go unnoticed until the final answer fails to check out against the original, unexpanded form, rather than against a line of working that might carry the same mistake forward unnoticed. This one habit, more than any formula, is what separates a confidently correct answer from a lucky one — and it costs only a few extra lines of arithmetic each time — a small, repeatable price for real certainty in the answer, rather than a guess that happens to look right on paper but was never actually confirmed against the original equation.

Why This Exercise Feels Harder Than It Is

Equations with variables on both sides often feel more difficult than the previous exercise's equations, even though the individual moves — transpose, expand a bracket, combine like terms — are all already familiar. The added difficulty is really just volume: more terms to track, more brackets to expand correctly, and more opportunities for a sign to slip somewhere along the way. Slowing down to do the two passes in order, rather than trying to transpose everything in one sweep, is what keeps that volume from turning into actual mistakes.

Bracket Discipline That Carries Forward

Handling variables on both sides, including through several layers of brackets, is exactly what the word problems in Exercise 2.4 require once they're translated into equations. The same bracket-expansion discipline is needed again for Exercise 2.5, where fractions get added into the mix as well. Looking further ahead, this exact skill of collecting like terms across an equals sign is what linear equations in two variables builds on in Class 9.