Linear Equations and Formula Rearrangement

Prerequisites: Section 1B Lessons 01–02, including brackets, common factors and domain restrictions. All practice questions in this lesson are original, not official TMUA questions.

What This Lesson Is About

A linear equation uses a variable to the first power and asks for values that make two expressions equal. Formula rearrangement uses the same operations to express one variable in terms of others. This lesson develops both skills while explaining when a division is justified.

We start with equations containing brackets and fractions. We then examine equations whose variable coefficient can be zero, make a chosen variable the subject of a formula, and use linear models in word problems. The same principle connects these topics: preserve the allowed solutions at every step.

1. Solving Linear Equations with Reversible Steps

A solution is an input that makes the original equality true. A reliable method transforms the equation without losing valid inputs or adding false ones. This is why the operation performed on both sides matters as much as the final number.

1.1 Equivalent equations: preserving the solution set

To solve an equation, find every allowed input that makes its two sides equal. Adding or subtracting the same expression on both sides preserves equality. Multiplying or dividing both sides by a known nonzero quantity is reversible; dividing by zero is never allowed.

An operation is reversible when it can be undone on the allowed inputs. Adding 3 can be undone by subtracting 3; multiplying by 2 can be undone by dividing by 2. Multiplying by zero cannot be reversed, because different starting values all become zero.

1.2 Brackets and fractions: organising the calculation

Expand brackets and clear constant denominators before collecting variable terms. When multiplying by a common denominator, multiply every term on both sides. Keep your final answer exact, and substitute it into the original equation.

2. Linear Equations with Parameters

Not every equation that looks linear determines one number. After collecting variable terms, the coefficient can disappear. We first classify that result, then use the classification when a coefficient contains a parameter.

2.1 The equation Ax = B: one, no or infinitely many solutions

After collecting terms, a linear equation has the form Ax = B, where A and B do not depend on x. If A ≠ 0, it has the unique solution B/A. If A = 0 and B ≠ 0, it is impossible. If A = B = 0, every allowed x works.

The three cases ask whether division by A is possible. If it is not, the remaining equality must be read as a statement about constants rather than as an instruction to find x.

2.2 Parameters: separating the zero-coefficient case

A parameter is a quantity treated as fixed while solving for another variable. Its value may determine which of the three cases occurs. Dividing by an expression in the parameter before considering its zeros can lose an entire case.

3. Rearranging a Formula to Make a Variable the Subject

To make a variable the subject means to write it alone on one side, expressed using the other quantities. When it appears in more than one term or in a denominator, collecting and factoring is often the key step.

3.1 Collecting the required variable before dividing

To make x the subject of a formula, first remove fractions on the allowed domain, then bring all terms containing x to one side. Factor out x. Only then divide by its coefficient, checking whether that coefficient can be zero.

3.2 Exceptional cases: returning to the original relation

Rearrangement changes the presentation of a relation; it does not remove restrictions. If a special parameter value makes the new denominator zero, return to the original relation to decide whether there are no solutions or many.

The values excluded by a rearranged denominator can represent different behaviours. The original formula tells you whether that case gives an impossible condition or leaves the requested variable free. Do not assume that division by zero always corresponds to no solutions.

4. Linear Models and the Requested Quantity

An equation becomes a model when its symbols represent quantities in a situation. Define those quantities before writing the equation, and interpret the algebraic result afterwards. Sometimes the question asks for an expression that can be read directly from a known equality.

4.1 Modelling a fixed amount and a per-unit amount

Choose a variable and define its units before writing a word problem as an equation. Translate fixed charges as constants and per-unit charges as coefficients. An algebraic solution must also satisfy the context, such as a nonnegative distance or an integer count.

4.2 Finding an expression without first finding the variable

Sometimes an expression can be found directly. If 3x − 2 = 7, then 12x − 8 is four times the known left side, so it equals 28 without first finding x. This works because the whole expression is multiplied by the same constant.

This shortcut still uses a reversible, justified relationship. Identify the multiple of the entire given expression; multiplying only its variable term would change the question.

5. Practice

The questions move from numerical coefficients to parameter cases and formula rearrangement. Finish by interpreting what the question actually requests. In every division, identify why the divisor is nonzero or treat its zero case separately.

1. Solve (x − 2)/5 + (x + 1)/2 = 3.

2. Classify 3(2x − 1) = 6x + 4.

3. Classify the solutions of (a − 1)x = 2a − 2 for real a.

4. For which real k does (k + 1)x = k − 1 have no solution?

5. Make h the subject of A = h(a + b)/2, assuming a + b ≠ 0.

6. Make x the subject of y = (3x − 1)/(x + 2), including excluded values.

7. If 5x + 2 = 9, find 15x + 6 without solving for x.

8. A service costs £7 plus £3 per whole item. Can a bill of £26 arise under this model? Explain.

Worked Solutions

1. Clear constant denominators and collect terms

Multiply by 10: 2(x − 2) + 5(x + 1) = 30. Then 7x + 1 = 30, so x = 29/7. Check: (x − 2)/5 = 3/7 and (x + 1)/2 = 18/7, summing to 3.

2. Recognise a contradictory constant statement

Expansion gives 6x − 3 = 6x + 4, or −3 = 4. There is no solution.

3. Separate the zero-parameter factor

For a ≠ 1, division by a − 1 gives x = 2. For a = 1, both sides are zero and every real x works. Direct substitution of x = 2 verifies the nonzero case.

4. Find when the variable coefficient vanishes

k = −1 makes the equation 0x = −2, which has no solution. Every other k gives the unique solution (k − 1)/(k + 1).

5. Isolate the subject using its nonzero coefficient

Multiply by 2 and divide by a + b: h = 2A/(a + b). Substitution into h(a + b)/2 returns A. The stated assumption makes division valid.

6. Preserve both original and rearranged restrictions

The original requires x ≠ −2. Rearrangement gives yx + 2y = 3x − 1, hence x(y − 3) = −1 − 2y. Thus x = (2y + 1)/(3 − y), with y ≠ 3. At y = 3 the original equation is impossible. The new formula never yields −2, since 2y + 1 = −6 + 2y is impossible.

7. Multiply the complete known expression

Multiply the whole given expression by 3: 15x + 6 = 27.

8. Check the integer requirement in the model

Let n be a nonnegative integer. Then 7 + 3n = 26 would require n = 19/3, not an integer. No such bill arises. The neighbouring permitted counts 6 and 7 give £25 and £28.

6. Summary

• Use reversible operations and check the original equation.
• Classify Ax = B by whether its coefficient and constant vanish.
• Factor the requested variable before the final division in a rearrangement.
• Carry parameter, domain and contextual restrictions into the answer.