Linear equations: solving step by step

A linear equation states that two linear expressions have the same value. Solving it means finding every value of the unknown that makes the statement true.

For example, the equation

$3x+5=20$

has solution $x=5$, because substituting $5$ makes both sides equal to $20$.

Linear equations occur throughout A-level Mathematics. They appear directly, but they also form individual steps inside coordinate geometry, trigonometry, calculus, statistics and mechanics. The essential habit is to preserve equality while simplifying the equation.

Prerequisites

You should already be able to:

• calculate accurately with negative numbers and fractions;
• collect like terms;
• expand a single bracket;
• substitute a value into an expression.

Review exact arithmetic or algebraic manipulation and factorisation if these steps are not yet secure.

Expressions, equations and solutions

An expression such as $4x-7$ has no equals sign. It can be simplified or evaluated, but it cannot be solved by itself.

An equation such as

$4x-7=13$

asserts equality. A solution is a value that makes this assertion true. Here $x=5$ is a solution because

$4(5)-7=13.$

The letter does not have to be $x$. Solving $7-2t=15$ for $t$ uses exactly the same reasoning.

The balance principle

Think of the equals sign as a balance. If two quantities are equal, applying the same valid operation to both preserves equality:

$a=b\quad\Longrightarrow\quad a+c=b+c$

and, for $c\ne0$,

$a=b\quad\Longrightarrow\quad \frac{a}{c}=\frac{b}{c}.$

This explains the familiar instruction to “do the same to both sides”. Terms do not literally jump across the equals sign. What is often called moving a term is shorthand for adding or subtracting that term on both sides.

For instance,

$$\begin{aligned} x+8&=19\\ x+8-8&=19-8\\ x&=11. \end{aligned}$$

Writing every elementary line is not always necessary, but the operation must remain logically valid.

One step and two step equations

To isolate the unknown, undo operations in reverse order.

Worked example 1: a positive coefficient

Solve $5x-7=28$.

The expression $5x-7$ means multiply by $5$, then subtract $7$. Undo subtraction first:

$$\begin{aligned} 5x-7&=28\\ 5x&=35\\ x&=7. \end{aligned}$$

Check:

$5(7)-7=35-7=28.$

Worked example 2: a negative coefficient

Solve $4-3x=19$.

$$\begin{aligned} 4-3x&=19\\ -3x&=15\\ x&=-5. \end{aligned}$$

Dividing $15$ by $-3$ gives $-5$. A negative solution is not evidence of an error.

Misconception: divide only one term

From $6x+9=27$, first obtain $6x=18$, then divide both sides by $6$.

Do not divide only $6x$ in the original equation. If you choose to divide the original equation by $3$, every term must be divided:

$2x+3=9.$

Self-check 1

Solve:

1. $7x+4=39$;
2. $12-5x=-8$;
3. $\dfrac{x}{4}+3=10$.

Answers

1. $7x=35$, so $x=5$.
2. $-5x=-20$, so $x=4$.
3. $\dfrac{x}{4}=7$, so $x=28$.

Unknowns on both sides

Collect the unknown terms on one side and constants on the other. Either side is valid, but choosing the side that leaves a positive coefficient often makes the arithmetic easier.

Worked example 3

Solve $7x+2=3x+18$.

Subtract $3x$ from both sides:

$$\begin{aligned} 7x+2&=3x+18\\ 4x+2&=18\\ 4x&=16\\ x&=4. \end{aligned}$$

Checking in the original equation gives

$7(4)+2=30,\qquad 3(4)+18=30.$

Worked example 4: decimals and a negative solution

Solve $1.6x-3=0.4x+6$.

$$\begin{aligned} 1.6x-3&=0.4x+6\\ 1.2x-3&=6\\ 1.2x&=9\\ x&=7.5. \end{aligned}$$

Decimals are permitted, but exact fractions are often safer when values recur or do not terminate.

Equations containing brackets

Expand every bracket carefully, then collect like terms before isolating the unknown.

Worked example 5: one bracket

Solve $4(2x-3)+5=25$.

$$\begin{aligned} 4(2x-3)+5&=25\\ 8x-12+5&=25\\ 8x-7&=25\\ 8x&=32\\ x&=4. \end{aligned}$$

The multiplier $4$ applies to both terms inside the bracket.

Worked example 6: brackets on both sides

Solve $3(2x+1)-4=2(1-x)+15$.

First expand and simplify each side:

$$\begin{aligned} 6x+3-4&=2-2x+15\\ 6x-1&=17-2x. \end{aligned}$$

Now collect the $x$ terms:

$$\begin{aligned} 8x-1&=17\\ 8x&=18\\ x&=\frac{18}{8}=\frac94. \end{aligned}$$

Keep the exact fraction unless a decimal is requested.

Misconception: a minus before a bracket changes one sign

In

$5x-(2x-7),$

the bracket is multiplied by $-1$, so both signs change:

$5x-2x+7=3x+7.$

It is not $5x-2x-7$.

Self-check 2

Solve:

1. $5(x-2)=3x+8$;
2. $2(3x+4)-5=4(x-1)$;
3. $7-2(3x+1)=4(x-5)$.

Answers

1. $5x-10=3x+8$, so $2x=18$ and $x=9$.
2. $6x+3=4x-4$, so $2x=-7$ and $x=-\dfrac72$.
3. $5-6x=4x-20$, so $25=10x$ and $x=\dfrac52$.

Equations containing fractions

Fractions can be handled in two reliable ways:

• isolate a single fraction, then multiply by its denominator;
• multiply every term by the lowest common multiple of all denominators.

The second method clears several fractions at once.

Worked example 7: one algebraic fraction

Solve

$\frac{3x-5}{4}=7.$

Multiply both sides by $4$:

$$\begin{aligned} 3x-5&=28\\ 3x&=33\\ x&=11. \end{aligned}$$

The fraction bar groups the entire numerator $3x-5$.

Worked example 8: clear two denominators

Solve

$\frac{x-1}{3}+\frac{x+2}{4}=5.$

The lowest common multiple of $3$ and $4$ is $12$. Multiply every term by $12$:

$$\begin{aligned} 4(x-1)+3(x+2)&=60\\ 4x-4+3x+6&=60\\ 7x+2&=60\\ 7x&=58\\ x&=\frac{58}{7}. \end{aligned}$$

Check without converting to a rounded decimal:

$$\frac{\frac{58}{7}-1}{3}+\frac{\frac{58}{7}+2}{4} =\frac{17}{7}+\frac{18}{7}=5.$$

Worked example 9: an unknown denominator is different

The equation

$\frac{5}{x}=2$

is not linear because the unknown is in a denominator. However, it can still be solved here by noting $x\ne0$ and multiplying by $x$:

$5=2x\quad\Longrightarrow\quad x=\frac52.$

Do not assume that every equation which eventually becomes $ax=b$ was linear in its original form. Equations with variable denominators require attention to excluded values. See algebraic fractions for the full method.

Self-check 3

Solve:

1. $\dfrac{2x+1}{5}=3$;
2. $\dfrac{x}{2}-\dfrac{x-1}{3}=4$;
3. $\dfrac{2x-3}{4}=\dfrac{x+5}{6}$.

Answers

1. $2x+1=15$, so $x=7$.
2. Multiply by $6$: $3x-2(x-1)=24$. Hence $x+2=24$ and $x=22$.
3. Multiply by $12$: $3(2x-3)=2(x+5)$. Hence $6x-9=2x+10$, so $x=\dfrac{19}{4}$.

One solution, no solution or infinitely many solutions

Most linear equations in one unknown have one solution, but cancellation can reveal two special cases.

Worked example 10: no solution

Solve $4(x+2)=4x+3$.

Expanding gives

$4x+8=4x+3.$

Subtracting $4x$ from both sides leaves

$8=3,$

which is false. No value of $x$ can make the original equation true, so there is no solution.

Worked example 11: infinitely many solutions

Solve $3(2x-1)=6x-3$.

Expanding gives

$6x-3=6x-3,$

which is true for every real $x$. The two sides are equivalent expressions, so there are infinitely many solutions.

Do not divide by the coefficient of $x$ after it has cancelled to zero. Statements such as $0x=5$ and $0x=0$ must be interpreted, not divided by zero.

Final statement	Meaning	Solution set
$ax=b$, where $a\ne0$	one value works	$x=\dfrac ba$
$0=c$, where $c\ne0$	contradiction	no solution
$0=0$	identity	every real value

Forming a linear equation

In a worded problem, define the unknown, translate each relationship, solve, then interpret the result in context.

Worked example 12: consecutive integers

Three consecutive integers have sum $96$. Find them.

Let the first integer be $n$. The next two are $n+1$ and $n+2$.

$$\begin{aligned} n+(n+1)+(n+2)&=96\\ 3n+3&=96\\ 3n&=93\\ n&=31. \end{aligned}$$

The integers are

$\boxed{31,32,33}.$

Their sum is $96$, so the answer satisfies the context.

Worked example 13: perimeter

A rectangle has length $2x+3$ cm and width $x-1$ cm. Its perimeter is $34$ cm. Find its dimensions.

Use $P=2l+2w$:

$$\begin{aligned} 2(2x+3)+2(x-1)&=34\\ 4x+6+2x-2&=34\\ 6x+4&=34\\ x&=5. \end{aligned}$$

Therefore the length is $2(5)+3=13$ cm and the width is $5-1=4$ cm. The perimeter check is $2(13)+2(4)=34$ cm.

Worked example 14: modelling a charge

A taxi fare is a fixed charge of £3.80 plus £1.60 per mile. A journey costs £15.80. Find the distance travelled.

Let the distance be $d$ miles:

$$\begin{aligned} 3.80+1.60d&=15.80\\ 1.60d&=12.00\\ d&=7.5. \end{aligned}$$

The journey was $\boxed{7.5\text{ miles}}$. The units belong in the final answer.

Checking and diagnosing an answer

Substitute your proposed solution into the original equation. Simplifying the original left and right sides independently is stronger than checking only your rearranged final line.

For $2(3x-1)=5x+7$ and proposed solution $x=9$:

$\text{LHS}=2(27-1)=52,$

$\text{RHS}=45+7=52.$

Since both sides agree, $x=9$ is correct.

A check can also expose the type of error:

• one side differs only by a sign, so inspect negative coefficients and brackets;
• one side differs by a constant factor, so inspect multiplication or division of every term;
• a decimal answer is close but not exact, so repeat the work with fractions;
• the algebra ends in a false constant statement, so the equation may genuinely have no solution.

Mixed self-check

1. Solve $9-4x=2x+21$.
2. Solve $5(2y-3)+4=3(y+2)$.
3. Solve $\dfrac{p+2}{3}-\dfrac{p-1}{2}=1$.
4. Classify the solutions of $2(3t+4)=6t+8$.
5. Classify the solutions of $7q-5=7(q+1)$.
6. The angles of a triangle are $x$, $2x+10^\circ$ and $3x-4^\circ$. Find all three angles.

Answers

1. $-6x=12$, so $x=-2$.
2. $10y-11=3y+6$, so $7y=17$ and $y=\dfrac{17}{7}$.
3. Multiply by $6$: $2(p+2)-3(p-1)=6$. Thus $2p+4-3p+3=6$, so $p=1$.
4. Both sides simplify to $6t+8$, so every real $t$ is a solution.
5. Simplification gives $-5=7$, so there is no solution.
6. Since angles in a triangle sum to $180^\circ$, $x+(2x+10)+(3x-4)=180.$ Hence $6x+6=180$, so $x=29^\circ$. The angles are $29^\circ$, $68^\circ$ and $83^\circ$.

What to learn next

Linear equations provide the algebraic engine for several connected topics:

• rearranging formulae generalises the same balance principle when several letters are present;
• simultaneous equations finds values satisfying two equations together;
• inequalities replaces equality with an order relation and introduces one crucial sign rule;
• straight-line graphs interprets linear relationships geometrically;
• quadratic equations develops methods for equations containing $x^2$.