2.4 - Forming and solving linear equations

Find an unknown while keeping both sides of an equation equal. You will handle brackets, unknowns on both sides and fractional coefficients. Then use the same reasoning to turn a situation into an equation and interpret its solution.

Balancing an equation

A taxi charges £4 plus £2.50 per kilometre. The total bill is £19. How far was the journey? We will form an equation for the distance and solve it. First, learn how to keep an equation balanced.

You already know

An expression such as $5x-7$ represents a value. An equation such as $5x-7=18$ says that two expressions have the same value. Inverse operations undo one another.

The two sides act like a balance. Add or subtract the same quantity on both sides. Multiply or divide both sides by the same nonzero number. Changing only one side usually destroys the equality.

Solution of an equation

A value of the unknown that makes the original equality true. For example, $x=5$ solves $5x-7=18$ , since $25-7=18$ . The value $x=4$ does not solve it, since $20-7=13$ .

A linear equation has no squared or higher powers of its unknown after simplification. The balancing method is not printed on the Foundation or Higher formulae sheet: learn to use inverse operations.

Worked example 1: reverse the operations

Question. Solve $5x-7=18$ .

Plan. Undo subtracting $7$ , then undo multiplication by $5$ .

\begin{aligned} 5x&=25 &&\text{add }7\text{ to both sides}\\ x&=5 &&\text{divide both sides by }5 \end{aligned}

Why add before dividing? It isolates the entire term $5x$ , leaving one operation to undo.

Answer. $x=5$ .

Check. In the original equation, $5(5)-7=25-7=18$ .

Change only the sign of the coefficient of $x$ . The same inverse operations now require division by a negative number.

Worked example 2: a negative coefficient

Question. Solve $-5x-7=18$ .

Plan. Add $7$ , then divide by $-5$ .

\begin{aligned} -5x&=25 &&\text{add }7\text{ to both sides}\\ x&=-5 &&\text{divide by }-5 \end{aligned}

Why retain the negative coefficient? It is the multiplier of $x$ , so its inverse is division by $-5$ , rather than by $5$ .

Answer. $x=-5$ .

Check. $-5(-5)-7=25-7=18$ .

Write one equivalent equation per line. Avoid joining unrelated results with equals signs.

Watch out

Wrong: $-5x=25$ gives $x=5$ . That divides by the wrong sign. Correct: $x=25\div(-5)=-5$ ; substitution exposes the mistake.

Now use a different negative coefficient and a different added constant.

Unknowns on both sides

If the unknown appears on both sides, collect its terms on one side and constants on the other. Expand brackets first when that makes the terms visible. The multiplier outside a bracket affects every term.

Either side can hold the unknown. Choosing the larger coefficient often avoids a negative coefficient, but both routes are valid. “Move a term and change its sign” is a shortcut for adding or subtracting the same term on both sides, not a separate rule.

Worked example 3: brackets and two sides

Question. Solve $4(x-2)=2x+10$ .

Plan. Expand, subtract $2x$ from both sides, then isolate $x$ .

\begin{aligned} 4x-8&=2x+10 &&\text{expand both terms}\\ 2x-8&=10 &&\text{subtract }2x\\ 2x&=18 &&\text{add }8\\ x&=9 &&\text{divide by }2 \end{aligned}

Why subtract $2x$ ? It gathers all unknown terms on the left without changing the solution.

Answer. $x=9$ .

Check. Left side: $4(9-2)=28$ . Right side: $2(9)+10=28$ .

Change only the coefficient on the right from $2$ to $3$ . The collection step leaves a different coefficient of $x$ .

Worked example 4: change one coefficient

Question. Solve $4(x-2)=3x+10$ .

Plan. Expand and subtract $3x$ , then remove the remaining constant.

\begin{aligned} 4x-8&=3x+10 &&\text{expand}\\ x-8&=10 &&\text{subtract }3x\\ x&=18 &&\text{add }8 \end{aligned}

Why does the answer double? The earlier equation needed $2x=18$ ; this one needs $x=18$ .

Answer. $x=18$ .

Check. $4(18-2)=64$ and $3(18)+10=64$ .

Not every equation has one solution. The equality $2(x+1)=2x+2$ simplifies to $2=2$ , so every real $x$ works. Changing just the last constant to $3$ gives $2=3$ , so no $x$ works. This small change turns infinitely many solutions into none. Do not divide by a zero coefficient when variable terms cancel.

Watch out

Wrong: $4(x-2)=4x-2$ . It multiplies only the first term. Correct: $4x-8$ . Expand the whole bracket before collecting terms.

Both expansion and collection are needed; neither method is supplied on either tier's formulae sheet. Change the right-hand coefficient once more.

Fractional coefficients

Clear constant denominators by multiplying every term on both sides by a common denominator. The lowest common multiple is convenient. Keep brackets around a numerator containing several terms.

The multiplier must reach whole-number terms too. This changes the presentation of the equation but preserves its solutions, since the multiplier is nonzero. Exact fraction arithmetic is an equally valid alternative. The denominator-clearing method is absent from both tier formulae sheets.

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Number

Algebra

Shape, space and measure

Handling data

2.4 - Forming and solving linear equations

Balancing an equation

Unknowns on both sides

Fractional coefficients

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