Secondary Mathematics: Equations, Balance and Checking

Contents · choose your starting point

1. 1. An equation makes a claim that a value can test
2. 2. What the balance principle does—and does not—mean
3. 3. Undo operations in the correct order
4. 4. Unknowns on both sides: collect relationships, not just terms
5. 5. Fractions are structure, not a reason to panic
6. 6. Decimal coefficients can be made easier to read
7. 7. Sometimes the correct answer is not a single number
8. 8. Write a model before solving a word problem
9. 9. Two unknowns need enough independent information
10. 10. Formula rearrangement solves for a different subject
11. 11. Variable denominators need a domain check first
12. 12. Extension: squaring can introduce candidates
13. 13. Extension: inequalities require an additional sign decision
14. 14. A capstone model: prices, totals and a decision
15. 15. Independent practice
16. 16. Worked answers
17. 17. Make checking a separate act
18. 18. Continue through the learning hub
19. Sources and scope

Secondary Mathematics · Worked Repair Guide 02

Solving an equation means finding the values that make its two sides equal. It does not mean moving symbols until the page looks familiar. A reliable solution preserves the original conditions, isolates the unknown through justified operations, and checks the proposed answer in the original equation.

This guide develops that process from a one-step equation to brackets, fractions, unknowns on both sides, simultaneous equations, formula rearrangement and selected extension cases. Its central habit is “Name the operation, preserve the solutions, check the original.” Every example makes the reason for the next line visible.

There are two different skills to learn. One is carrying out algebra accurately. The other is knowing whether an operation is allowed and whether it can lose or introduce solutions. The first skill makes working fluent. The second makes the conclusion trustworthy.

The examples are original and illustrative. This is a worked learning companion, not a complete syllabus, an official examination paper or a promise of a particular grade. Use the core sections first; attempt extensions when they match the mathematics you are studying.

1. An equation makes a claim that a value can test

An expression such as 3x + 5 describes a quantity. An equation such as 3x + 5 = 20 claims that two quantities are equal. A solution is a permitted value of x that makes that claim true.

For x = 5, the left side is 3(5) + 5 = 20, matching the right side. For x = 4, the left side is 17, so 4 is not a solution. This simple test should remain available even when the equation becomes more complicated.

The equals sign is a statement of equality, not a signal meaning “and then”. A chain such as 3x + 5 = 20 − 5 = 15 contains an unintended claim that 3x + 5 equals 15. Instead, write separate valid equations:

3x + 5 = 20
3x = 15
x = 5.

Each line now says something true about the solution. The transformation between lines can be described: subtract 5 from both sides, then divide both sides by 3.

Begin with four entry tasks. Solve 4x − 7 = 13; solve 2(x + 3) = x + 11; decide whether 3x + 2 = 3x + 7 has a solution; and solve (x − 1)/3 = 4. The answers are 5, 5, no solution, and 13 respectively. Record which transformation caused uncertainty rather than using the total alone to choose your next step.

If calculation fails before you can explain the operation, return to Signed Numbers, Brackets and Algebraic Structure. If the algebra is sound but a word problem cannot be started, focus on the modelling sections here rather than repeating only symbolic equations.

2. What the balance principle does—and does not—mean

If two real quantities are equal, adding the same quantity to both preserves their equality. Subtracting the same quantity also preserves it. Multiplying or dividing both sides by the same non-zero number preserves the solution set of an equation.

The phrase “do the same thing to both sides” needs that qualification. Multiplying by zero makes both sides zero whatever they originally were. The equation 2x = 6 becomes 0 = 0, which no longer identifies x = 3. A valid solution has lost its information.

Similarly, dividing by an expression that might be zero requires care. From x(x − 4) = 0, division by x would remove the solution x = 0. You may divide in a separately declared case where x ≠ 0, but you must still examine the omitted case.

The balance picture is a useful introduction, but the deeper criterion is whether the next equation has exactly the same permitted solutions. For ordinary linear equations, adding or subtracting the same expression is safe, and division by a known non-zero coefficient is safe. For other operations, the direction and conditions need inspection.

Worked example: solve 5x + 8 = 33. Subtract 8 from both sides to get 5x = 25. Divide both sides by 5 to get x = 5. The coefficient 5 is non-zero, so the division is reversible. Substituting into the original gives 25 + 8 = 33.

The number 8 did not physically move across the equals sign and change its character. Subtracting 8 from both sides created the new equation. The familiar “move and change sign” shortcut is only safe when you can reconstruct the operation it abbreviates.

3. Undo operations in the correct order

An equation can describe a sequence of operations performed on an unknown. In 4x − 7 = 13, x was multiplied by 4 and then 7 was subtracted. To recover x, undo the outer operation first: add 7, then divide by 4.

4x − 7 = 13
4x = 20
x = 5.

Check: 4(5) − 7 = 13. The inverse operations return us to the original quantity.

Now compare 4(x − 7) = 12. Here the unknown was reduced by 7 before the whole result was multiplied by 4. Undo the multiplication first:

x − 7 = 3
x = 10.

Expanding first would also work: 4x − 28 = 12 leads to 4x = 40 and x = 10. The best route is the one whose operations you can carry out and justify reliably. “Always expand every bracket first” is not necessary.

Negative coefficient example: solve −3x + 4 = 19. Subtract 4 to get −3x = 15. Divide by −3 to get x = −5. The negative coefficient must remain part of the divisor. The original check gives −3(−5) + 4 = 19.

A useful prediction accompanies the algebra: −3x has to be positive 15, so x must be negative. This does not replace the calculation, but it can reject a positive answer before it reaches the final line.

4. Unknowns on both sides: collect relationships, not just terms

When x appears on both sides, choose a lawful subtraction that reduces the number of places where x occurs. For example:

7x − 9 = 3x + 15.

Subtract 3x from both sides to obtain 4x − 9 = 15. Add 9, giving 4x = 24. Divide by 4, giving x = 6. The check is 42 − 9 = 18 + 15 = 33.

Subtracting 7x first is equally valid. It produces −9 = −4x + 15, then −24 = −4x, then x = 6. A learner should not reject a correct route merely because it differs from a memorised layout.

Bracket example: solve 3(2x − 1) = 4(x + 2) + 5. Expanding accurately gives 6x − 3 = 4x + 13. Subtract 4x and add 3 to obtain 2x = 16, so x = 8. In the original, the left side is 3(15) = 45, and the right side is 4(10) + 5 = 45.

The equation involves two forms of preservation. Within one side, distribution changes an expression into an equal expression. Between the two sides, subtracting the same quantity preserves the equality. Mixing those jobs can produce errors such as changing a sign merely because a term is being regrouped on the same side.

Write an operation beside a difficult line: “expand”, “subtract 4x from both sides”, “add 3 to both sides”. These annotations can be removed when the reasoning is secure. Their purpose is to expose the decision, not to make every solution unnecessarily long.

5. Fractions are structure, not a reason to panic

For an equation containing numerical denominators, multiply every term on both sides by a suitable common multiple. This clears the denominators without changing the solutions because that multiplier is a known non-zero number.

Worked example: solve (x − 2)/3 + (x + 1)/2 = 4. The least common multiple of 3 and 2 is 6. Multiplying the entire equation by 6 gives:

2(x − 2) + 3(x + 1) = 24
2x − 4 + 3x + 3 = 24
5x − 1 = 24
x = 5.

Check in the original: (5 − 2)/3 + (5 + 1)/2 = 1 + 3 = 4. A check only in the cleared equation would not reveal a mistake in how the denominators were cleared.

The fraction bar groups each numerator. After multiplying (x − 2)/3 by 6, the result is 2(x − 2), not 2x − 2. The multiplier reaches the complete numerator. This is the same distributive structure taught in the first guide.

Second example: solve (2x + 1)/4 − (x − 3)/6 = 2. Multiplying every term by 12 gives 3(2x + 1) − 2(x − 3) = 24. Expanding gives 6x + 3 − 2x + 6 = 24, hence 4x = 15 and x = 15/4.

That fractional solution is exact and entirely acceptable. In the original, the fractions are 17/8 and 1/8, whose difference is 2. Do not round an exact intermediate answer simply because a whole number would feel more comfortable.

6. Decimal coefficients can be made easier to read

An equation such as 0.4x + 1.2 = 3.8 can be solved directly, but multiplying every term by 10 produces 4x + 12 = 38. Subtract 12 and divide by 4 to obtain x = 26/4 = 6.5.

The multiplication changes the scale of both sides, not the set of solutions. The original check is 0.4(6.5) + 1.2 = 2.6 + 1.2 = 3.8.

Be clear whether the decimal numbers are exact values in the question or rounded measurements from a real setting. In a pure equation exercise, a written terminating decimal is normally treated as the stated exact number. In a measurement model, the final answer may need an appropriate degree of precision. The context determines that interpretation; algebra alone does not.

A similar option is to use fractions. Since 0.4 = 2/5 and 1.2 = 6/5, multiplying the fractional equation by 5 also works. There is no need to switch representations unless doing so makes the calculation more transparent.

Repair contrast: 0.2(x + 3) = 1.4 becomes 2(x + 3) = 14 after multiplying by 10. It does not become 2x + 3 = 14. Clearing decimals does not suspend the bracket. The solution is x + 3 = 7, so x = 4.

7. Sometimes the correct answer is not a single number

A linear-looking equation can have one solution, no solution or every real number as a solution. Its final simplified statement tells us which situation we have.

Consider 2(x + 3) = 2x + 6. Expanding gives 2x + 6 = 2x + 6. Subtracting 2x and then 6 gives 0 = 0. Every real x satisfies the original equation. This is an identity on the real numbers.

Now consider 2(x + 3) = 2x + 9. Expansion gives 2x + 6 = 2x + 9, hence 6 = 9. That statement is false, so there is no solution. We have not failed to finish the equation. We have shown that no value can satisfy it.

Do not turn 0x = 3 into x = 3/0. The conclusion is that zero times any real number cannot equal 3. Likewise 0x = 0 does not yield a unique value x = 0; it is true for every real x.

More generally, ax + b = cx + d becomes (a − c)x = d − b. If a − c is non-zero, there is the single solution x = (d − b)/(a − c). If a = c, inspect b and d. Equal constants give all real solutions; unequal constants give none.

For a learner, this section repairs an important expectation: “solve” does not guarantee one tidy numeral. Read the mathematical conclusion that the equation actually supports.

8. Write a model before solving a word problem

A word problem has an additional stage that a printed equation has already supplied: constructing the relationship. Start by identifying the unknown, its unit or meaning, and the condition connecting it to the known information.

Original example: a club buys three identical notebooks and a $2 folder for a total of $17. Let n be the price of one notebook in dollars. The model is 3n + 2 = 17. Solving gives n = 5. The context check is three $5 notebooks plus a $2 folder, totalling $17.

The variable represents one notebook’s price, not the total notebook expenditure. A correct equation followed by “the notebooks cost $5 altogether” misinterprets the variable. Defining it at the start protects the final sentence.

Comparison example: one number is 7 more than another, and their sum is 31. Let the smaller number be x; the larger is x + 7. Then x + (x + 7) = 31, giving 2x = 24 and x = 12. The numbers are 12 and 19. Check both conditions: the difference is 7 and the sum is 31.

Geometry example: a rectangle has width w centimetres and length w + 4 centimetres. Its perimeter is 36 centimetres. The model is 2w + 2(w + 4) = 36. Solving gives 4w + 8 = 36, so w = 7 and the length is 11. Check: 2(7 + 11) = 36.

The requirement w > 0 comes from the stated rectangle. An equation can sometimes yield a numerical candidate that a context rejects. Always carry the meaning of the unknown through to the final answer.

9. Two unknowns need enough independent information

One equation such as x + y = 10 usually does not determine both x and y uniquely over the real numbers. The pairs (2, 8), (4, 6) and (10, 0) all satisfy it. A second independent relationship can select a single pair.

Worked example: suppose x + y = 10 and 2x + y = 16. Subtract the first equation from the second. The y terms cancel, leaving x = 6. Substitution into x + y = 10 gives y = 4. Check both original equations, not merely the one used last.

This method is elimination. It works because subtracting equal quantities from equal quantities preserves equality. The cancellation is planned by choosing matching coefficients; it is not a visual trick involving nearby symbols.

A scaled example: solve 2x + 3y = 19 and 3x − 2y = 9. Multiply the first equation by 2 and the second by 3:

4x + 6y = 38
9x − 6y = 27.

Adding gives 13x = 65, so x = 5. The first original equation then gives 10 + 3y = 19, so y = 3. The second check is 15 − 6 = 9.

Two printed equations are not automatically two independent pieces of information. The pair x + y = 10 and 2x + 2y = 20 describes the same relationship twice. By contrast, x + y = 10 and 2x + 2y = 25 is inconsistent. Later, the graph guide interprets these as coincident lines and parallel distinct lines.

10. Formula rearrangement solves for a different subject

A formula relates quantities. Rearranging it means isolating a chosen quantity while preserving the relationship and its conditions. Start by locating every occurrence of that quantity.

For P = 2l + 2w, make w the subject. Subtract 2l, then divide by 2:

P − 2l = 2w
w = (P − 2l)/2 = P/2 − l.

The denominator applies to the whole numerator. The expression P − l is not equivalent to (P − 2l)/2.

A fraction example: given v = d/t with t ≠ 0, make t the subject. Multiplying by t gives vt = d. If v ≠ 0, division by v gives t = d/v, while retaining any original contextual restrictions. If v = 0, that rearranged fraction is unavailable; the original equation requires d = 0 and does not uniquely determine a non-zero t.

Repeated subject example: make x the subject of y = (2x + 3)/(x − 1). The original restriction is x ≠ 1. Multiply by x − 1:

yx − y = 2x + 3
yx − 2x = y + 3
x(y − 2) = y + 3.

Hence x = (y + 3)/(y − 2), with y ≠ 2. The value y = 2 is impossible in the original relation: it would lead to 2x − 2 = 2x + 3. For permitted y, the resulting x cannot equal 1, since that would require 3 = −2.

This extension demonstrates why collecting the subject into one factor matters. A variable appearing twice cannot be isolated by undoing only one of its appearances.

11. Variable denominators need a domain check first

Before clearing a denominator containing x, write which values would make that denominator zero. Those inputs are not candidates for the original equation.

Worked example: solve (x + 1)/(x − 2) = 2. First state x ≠ 2. Multiply by the permitted non-zero denominator to get x + 1 = 2x − 4, hence x = 5. The original check gives 6/3 = 2, so the candidate is valid.

A revealing contrast: solve (x² − 4)/(x − 2) = 4. The restriction is again x ≠ 2. Factorisation simplifies the left side to x + 2 on that domain. Solving x + 2 = 4 produces x = 2, but the original denominator is zero there. Therefore the original equation has no solution.

It is not enough to solve the simplified equation and stop. The original domain still controls which candidates can be admitted. This is one reason for keeping the first equation visible while checking.

For ordinary linear equations with only numerical non-zero denominators, these extra restrictions do not arise. Do not burden every elementary equation with unnecessary machinery. Use the domain check when an operation or expression actually creates the issue.

A compact working format is “Restriction → transformation → candidate → original check”. It separates the four jobs clearly and prevents the candidate from being announced as a solution before it has passed the relevant conditions.

12. Extension: squaring can introduce candidates

Squaring both sides of an equation preserves a true equality, but it can remove information about signs. From a = b, we can conclude a² = b². The reverse direction does not always hold, because 3² = (−3)² although 3 ≠ −3.

Consider √(x + 6) = x, where the square root denotes the non-negative square root. The original equation requires x ≥ 0. Squaring gives x + 6 = x², or x² − x − 6 = 0. Factorising gives (x − 3)(x + 2) = 0, producing the candidates 3 and −2.

Check them in the original. At x = 3, √9 = 3, so 3 is a solution. At x = −2, √4 = 2, not −2, so −2 is rejected. The squaring step produced a necessary condition, not an automatically equivalent equation on all real inputs.

Another familiar example is x² = 16, whose real solutions are x = 4 and x = −4. Writing only x = 4 loses a solution. But √16 is the single number 4, not “±4”. Distinguish solving an equation from evaluating the principal square root symbol.

The lesson is not to avoid squaring. It is to know what information the operation can discard and to check accordingly. The same habit appears in more advanced mathematics whenever a transformation has conditions or is not reversible.

13. Extension: inequalities require an additional sign decision

Equations and inequalities share many algebraic operations, but multiplying or dividing an inequality by a negative number reverses its direction. For example, −2x < 8 gives x > −4 after division by −2.

Why reverse the sign? Since 2 < 5, multiplying both numbers by −1 gives −2 > −5. Negation reverses their order on the number line. This is a statement about order, not about the equals sign.

Worked example: solve 5 − 3x ≥ 14. Subtract 5 to get −3x ≥ 9. Divide by −3 and reverse the direction, giving x ≤ −3. Check a permitted boundary value: at x = −3, the left side is 14. Check an interior value such as −4: the left side is 17, which is at least 14. A value such as 0 fails.

Do not mechanically transfer this sign reversal to equations. Dividing −3x = 9 by −3 gives x = −3; the equality does not acquire a direction. Similarly, multiplying an inequality by an expression whose sign is unknown requires cases. The expression may be positive, negative or zero, and those cases do not obey one uniform step.

This section is included as a boundary comparison. It shows how two similar-looking tasks can require different conditions. Use it when inequalities are part of the learner’s current course, rather than treating it as a prerequisite for the first linear equations.

14. A capstone model: prices, totals and a decision

An imaginary school event sells adult tickets at $8 and student tickets at $5. It sells 70 tickets and collects $440. How many tickets of each type were sold?

Let a be the number of adult tickets and s the number of student tickets. The two relationships are:

a + s = 70
8a + 5s = 440.

Substitute s = 70 − a into the revenue equation:

8a + 5(70 − a) = 440
8a + 350 − 5a = 440
3a = 90
a = 30.

Then s = 40. Both counts are non-negative integers, as the context requires. The ticket check is 30 + 40 = 70. The revenue check is 30 × 8 + 40 × 5 = 240 + 200 = 440.

There is also a useful baseline method. If all 70 tickets cost $5, revenue would be $350. Each adult ticket adds $3 above that baseline. The extra $90 therefore represents 30 adult tickets. This is the same linear relationship expressed verbally: 350 + 3a = 440.

Now change the collected total to $441 while keeping all other assumptions. Algebra gives a = 91/3, not a whole number. Under the exact stated prices and counts, that data set has no feasible ticket allocation. Do not round to 30 and claim the original total was met. A non-integer candidate is evidence that the stated model and data do not fit together.

This is what checking in context adds beyond symbolic accuracy. Mathematics can reveal an inconsistent scenario, not merely calculate a number to place in an answer box.

15. Independent practice

Write the operation that justifies each difficult transition. For contextual questions, define the variable and check every stated condition. The final four tasks are extensions.

1. Solve 6x + 5 = 29.
2. Solve −4x + 7 = 23.
3. Solve 3(x − 2) = 21.
4. Solve 5x − 8 = 2x + 13.
5. Solve 2(3x + 4) = 5x − 7.
6. Solve (x + 2)/4 = 3.
7. Solve (x − 1)/2 + (x + 3)/3 = 6.
8. Solve 0.3x + 0.8 = 2.9.
9. Determine the solution set of 4(x − 2) = 4x − 8.
10. Determine the solution set of 5x + 1 = 5x − 6.
11. Two numbers sum to 46. The larger is 10 more than the smaller. Find both.
12. A rectangle has length 3 cm more than its width and perimeter 30 cm. Find its dimensions.
13. Solve x + y = 12 and 2x − y = 3.
14. Make h the subject of A = (1/2)bh, explaining any required divisor condition.
15. Solve (x + 4)/(x − 1) = 3, with its restriction.
16. Solve x(x − 7) = 0 without losing a solution.
17. Solve √(x + 12) = x over the real numbers.
18. Solve 8 − 2x > 14.

16. Worked answers

1. x = 4. Subtract 5 to obtain 6x = 24, then divide by 6. The original left side is 24 + 5 = 29.

2. x = −4. Subtract 7 to obtain −4x = 16, then divide by −4. Check −4(−4) + 7 = 23. The sign of the divisor is part of the operation.

3. x = 9. Divide both sides by 3 to get x − 2 = 7, then add 2. Expansion gives an alternative route, 3x − 6 = 21, with the same solution.

4. x = 7. Subtract 2x and add 8 to obtain 3x = 21. Both sides of the original then equal 27.

5. x = −15. Expand to 6x + 8 = 5x − 7. Subtracting 5x and 8 yields x = −15. The original left side is 2[3(−15) + 4] = 2(−41) = −82. The right side is 5(−15) − 7 = −82. Evaluate both sides separately so that an error in the check does not conceal a correct solution.

6. x = 10. Multiply the whole equation by 4 to obtain x + 2 = 12. Subtract 2. The numerator is one grouped object.

7. x = 33/5. Multiply by 6: 3(x − 1) + 2(x + 3) = 36. This gives 5x + 3 = 36, hence x = 33/5. In the original, the two contributions are 14/5 and 16/5, whose sum is 6.

8. x = 7. Multiplying by 10 gives 3x + 8 = 29. Therefore 3x = 21. The original check is 2.1 + 0.8 = 2.9.

9. Every real number. Expanding yields the same expression on both sides. Do not mistake the identity 0 = 0 for a conclusion that x must equal zero.

10. No solution. Subtracting 5x yields 1 = −6, which is false regardless of x. Division is neither needed nor available to repair the contradiction.

11. Eighteen and twenty-eight. Let the smaller be x. Then x + x + 10 = 46, so x = 18. Both the sum and the difference must be checked.

12. Width 6 cm; length 9 cm. The model 2w + 2(w + 3) = 30 gives 4w = 24. The dimensions are positive and their perimeter is 30 cm.

13. x = 5 and y = 7. Add the equations to obtain 3x = 15. Then substitute into x + y = 12. The second original equation gives 10 − 7 = 3.

14. h = 2A/b, for b ≠ 0. Multiply by 2 and divide by b. In a triangle context, a positive base supplies that condition. If b = 0 in the bare algebraic relation, the formula does not determine h by division.

15. x = 7/2, with x ≠ 1. Multiply by x − 1 within the allowed domain: x + 4 = 3x − 3. Thus 2x = 7. The original fraction is (15/2)/(5/2) = 3.

16. x = 0 or x = 7. A product of real numbers is zero only if at least one factor is zero. Set x = 0 or x − 7 = 0. Dividing by x at the outset would discard the first solution.

17. x = 4 only. The original requires x ≥ 0. Squaring gives x² − x − 12 = 0, or (x − 4)(x + 3) = 0. The candidate −3 fails the original sign requirement; 4 gives √16 = 4.

18. x < −3. Subtract 8 to obtain −2x > 6. Division by −2 reverses the direction. At x = −4, the original left side is 16, which is greater than 14; the boundary −3 is excluded.

17. Make checking a separate act

A check should begin with the original question, not with the last line of the solution. Copy the proposed value into the original left and right sides and evaluate them independently. For a model, also check the variable’s meaning, the units and any feasibility conditions such as whole-number counts.

For a linear equation solved by reversible steps, substitution confirms the candidate’s membership in the original solution set. For an equation transformed by squaring, or by division involving a variable, checking also helps reject invalid candidates. However, checking found candidates alone does not prove that no other solutions were lost. The transformation chain must still be examined for omissions.

This distinction protects against two different failures. An introduced solution passes through a broader transformed equation but fails the original. A lost solution never reaches the checking stage. To avoid both, keep conditions attached to operations and inspect any step that divides by a variable expression.

When revising, classify the first error as reading, transformation, arithmetic, domain, or contextual interpretation. Then choose a fresh question that tests that particular issue. There is no need to complete twenty equations with simple integer coefficients if the repeated problem is an undefined denominator in a different kind of equation.

A suggested independent check session contains one equation from a repaired type, one from an earlier secure type and one unfamiliar mixed question. Record whether a hint was needed. Treat the result as evidence about that session, not a permanent label attached to the learner.

18. Continue through the learning hub

Return to the BTT Mathematics Hub for the wider subject library. The BTT Mathematical Lab is the diagnostic route when a repeated difficulty needs closer investigation.

Within this batch, Signed Numbers, Brackets and Algebraic Structure repairs the expressions inside an equation. Ratio, Percentage and the Correct Base shows how to choose the right quantity before forming a relationship. Graphs, Tables and Relationships makes an equation visible as a set of coordinate pairs and explains why two lines can have one, none or infinitely many common solutions.

Sources and scope

The mathematical explanations are supported by the displayed derivations, explicit conditions and original checks. The sample models are invented for teaching; ticket prices and purchases are not statements about an actual event or business.

MOE’s secondary curriculum doorway should be used to identify the relevant school and subject-level framework. Not every extension in this guide is intended for every secondary year or subject level. Framework checked on 6 September 2026. The proposed practice routines are editorial suggestions, not an official assessment method or a tested guarantee of improvement.