1.2.2 Solve linear equations

What you'll learn

How to recognise a linear equation in one unknown.
How to preserve equality while rearranging and simplifying.
How to solve equations involving brackets, fractions and unknowns on both sides.
How to form and solve linear equations in modelling problems.

What is a linear equation?

An equation is a mathematical statement saying that two expressions are equal. The expression on the left of the equals sign has the same value as the expression on the right.

An unknown is a quantity whose value you need to find. It is usually represented by a letter such as $x$ .

A solution is a value of the unknown that makes the equation true.

Definition

Linear equation in one unknown

A linear equation in one unknown can be rearranged into the form $a x + b = 0$ , where $a$ and $b$ are constants and $a≠0a\neq 0$ .

The word linear means that the unknown has power 1. For example, $3 x + 7 = 19$ is linear. Equations containing $x^2$ , $x\sqrt{x}$ or $1x\frac{1}{x}$ are not linear equations.

Key Idea

The aim

Solve an equation by isolating the unknown: rearrange it until the unknown is alone on one side of the equals sign.

Equality and balancing

You can think of an equation as a balanced pair of expressions. To preserve that balance, you must perform the same valid operation on both sides.

For example, you may:

add the same quantity to both sides;
subtract the same quantity from both sides;
multiply both sides by the same non-zero quantity;
divide both sides by the same non-zero quantity.

These operations produce an equivalent equation, meaning an equation with the same solution.

Solving a two-step equation

Example

Solving a two-step equation

Solve $5 x - 7 = 18$ .

Add 7 to both sides to remove the constant term from the left:
$5 x - 7 + 7 = 18 + 7$
This simplifies to $5 x = 25$ .
Divide both sides by 5 to make the coefficient of $x$ equal to 1:
$5x5=255\frac{5x}{5}=\frac{25}{5}$
Therefore, $x = 5$ .
Check the solution in the original equation:
$5 (5) - 7 = 25 - 7 = 18$
The left-hand side equals the right-hand side, so the solution is correct.

A coefficient is a number multiplying a variable. In $5 x$ , the coefficient of $x$ is 5.

Common Mistake

Changing signs without a reason

Do not simply “move” a term and change its sign. For instance, going from $5 x - 7 = 18$ to $5 x = 18 + 7$ is shorthand for adding 7 to both sides. Thinking in terms of balanced operations reduces sign errors.

Collecting unknown terms

When the unknown occurs on both sides, collect all terms involving the unknown on one side and all constant terms on the other.

You may choose either side for the unknown. It is often convenient to choose the side that gives a positive coefficient.

Example

Collecting unknowns on one side

Solve $7 x - 4 = 3 x + 20$ .

Subtract $3 x$ from both sides to collect the unknown terms on the left:
$7 x - 3 x - 4 = 20$
This gives $4 x - 4 = 20$ .
Add 4 to both sides:
$4 x = 24$
Divide both sides by 4:
$x = 6$
Check using both sides of the original equation:
$7 (6) - 4 = 38$
and
$3 (6) + 20 = 38.$
Both sides have the same value, confirming the solution.

Tip

Keep the coefficient positive

If possible, collect the unknown terms on the side with the larger coefficient. This often avoids working with a negative coefficient, although either method is valid.

Equations containing brackets

Before collecting terms, remove brackets by expanding them. Expanding means multiplying every term inside a bracket by the factor outside it.

For example:

3 (x - 4) = 3 x - 12

If there is a minus sign before a bracket, every sign inside the bracket changes when it is expanded:

- (2 x - 5) = - 2 x + 5

Expanding before solving

Example

Solving an equation with brackets

Solve $4 (2 x - 3) = 3 (x + 6)$ .

Expand both brackets:
$8 x - 12 = 3 x + 18$
Subtract $3 x$ from both sides:
$5 x - 12 = 18$
Add 12 to both sides:
$5 x = 30$
Divide both sides by 5:
$x = 6$
Substitute $x = 6$ into the original equation. The left-hand side is $4 (12 - 3) = 36$ , and the right-hand side is $3 (6 + 6) = 36$ , so the solution is correct.

Common Mistake

Incomplete expansion

In $4 (2 x - 3)$ , the 4 multiplies both terms, giving $8 x - 12$ . Writing $8 x - 3$ leaves the second term unmultiplied.

Equations involving fractions

A linear equation may contain fractional coefficients or algebraic fractions. You can often simplify the working by multiplying every term by a common multiple of the denominators.

The lowest common multiple, or LCM, is the smallest positive number divisible by every denominator.

Example

Clearing fractional coefficients

Solve

x−13+x+22=6.\frac{x-1}{3}+\frac{x+2}{2}=6.

The LCM of the denominators 3 and 2 is 6. Multiply every term on both sides by 6:
$6(x−13)+6(x+22)=6(6).6\left(\frac{x-1}{3}\right)+6\left(\frac{x+2}{2}\right)=6(6).$
Cancel the denominators and simplify:
$2 (x - 1) + 3 (x + 2) = 36.$
Expand and collect like terms:
$2x−2+3x+6=365x+4=36.\begin{aligned} 2x-2+3x+6&=36\\ 5x+4&=36. \end{aligned}$
Subtract 4 and then divide by 5:
$\Rightarrow x=\frac{32}{5}.$
Substitution into the original equation gives
$325−13+325+22=95+215=6,\frac{\frac{32}{5}-1}{3}+\frac{\frac{32}{5}+2}{2} =\frac{9}{5}+\frac{21}{5} =6,$
so the exact solution is $x=325x=\frac{32}{5}$ .

Common Mistake

Multiplying only part of an equation

When clearing fractions, multiply every term on both sides by the LCM. Brackets help to show that an entire numerator, such as $x - 1$ , is being treated as one expression.

Special outcomes

Most linear equations have exactly one solution, but simplifying may sometimes eliminate the unknown completely.

No solution

If simplification produces a false statement, such as $0 = 5$ , no value of the unknown can satisfy the original equation.

Infinitely many solutions

If simplification produces an identity such as $0 = 0$ , every permitted value of the unknown satisfies the original equation. An identity is an equality that is true for all values in its domain.

Example

Recognising a special outcome

Consider $3 (x + 2) = 3 x + 8$ .

Expand the left-hand side:
$3 x + 6 = 3 x + 8.$
Subtract $3 x$ from both sides:
$6 = 8.$
This statement is false, so the equation has no solution. The two sides always differ by 2, regardless of the value of $x$ .

Common Mistake

Do not force a numerical answer

If all the unknown terms cancel, interpret the remaining statement. A false statement means no solution; a statement that is always true means infinitely many solutions.

Forming an equation from a context

In a modelling problem, the equation may not be given to you. You must translate the information into algebra.

A good model should:

define the unknown clearly;
express the relevant quantities in terms of that unknown;
use the relationship stated in the problem;
produce an answer that is reasonable in context.

Example

Modelling a ticket purchase

Adult tickets cost £8 and student tickets cost £5. A group buys 14 tickets for £88. Find the number of student tickets.

Let $x$ be the number of student tickets. Since there are 14 tickets altogether, the number of adult tickets is $14 - x$ .
Form an equation for the total cost:
$5 x + 8 (14 - x) = 88.$
Expand and simplify:
$5x+112−8x=88112−3x=88.\begin{aligned} 5x+112-8x&=88\\ 112-3x&=88. \end{aligned}$
Subtract 112 from both sides, then divide by negative 3:
$\Rightarrow x=8.$
The group therefore buys 8 student tickets and 6 adult tickets. Checking the cost gives £40 plus £48, which is £88.

Key Idea

Interpret the solution

A numerical solution is not the end of a modelling problem. State what the value represents and check that it satisfies practical restrictions, such as being a non-negative whole number when counting people or objects.

A reliable solving routine

For a complicated linear equation, use this order:

Expand any brackets.
Clear fractions if this makes the equation simpler.
Collect like terms on each side.
Collect unknown terms on one side.
Collect constants on the other side.
Divide by the coefficient of the unknown.
Check by substitution into the original equation.

Exam technique

In the exam

Show balanced algebraic steps clearly; unexplained jumps make sign and arithmetic errors harder to detect.
Keep answers exact unless a decimal is requested, so write $325\frac{32}{5}$ rather than an unnecessarily rounded value.
Substitute your result into the original equation, especially when the equation contains negatives, brackets or fractions.
In a modelling question, define the unknown, form the equation and finish with a sentence interpreting the solution in context.

Self review

Check yourself

Can you solve $5 (2 x - 1) - 3 = 4 x + 16$ and verify your result?
What conclusion should you reach if an equation simplifies to $0 = - 7$ ?
How would you remove the denominators from $x4−x−36=2\frac{x}{4}-\frac{x-3}{6}=2$ ?

What you'll learn

How to recognise a linear equation in one unknown.
How to preserve equality while rearranging and simplifying.
How to solve equations involving brackets, fractions and unknowns on both sides.
How to form and solve linear equations in modelling problems.

What is a linear equation?

An equation is a mathematical statement saying that two expressions are equal. The expression on the left of the equals sign has the same value as the expression on the right.

An unknown is a quantity whose value you need to find. It is usually represented by a letter such as $x$ .

A solution is a value of the unknown that makes the equation true.

Definition

Linear equation in one unknown

A linear equation in one unknown can be rearranged into the form $a x + b = 0$ , where $a$ and $b$ are constants and $a≠0a\neq 0$ .

The word linear means that the unknown has power 1. For example, $3 x + 7 = 19$ is linear. Equations containing $x^2$ , $x\sqrt{x}$ or $1x\frac{1}{x}$ are not linear equations.

Key Idea

The aim

Solve an equation by isolating the unknown: rearrange it until the unknown is alone on one side of the equals sign.

Equality and balancing

You can think of an equation as a balanced pair of expressions. To preserve that balance, you must perform the same valid operation on both sides.

For example, you may:

add the same quantity to both sides;
subtract the same quantity from both sides;
multiply both sides by the same non-zero quantity;
divide both sides by the same non-zero quantity.

These operations produce an equivalent equation, meaning an equation with the same solution.

Solving a two-step equation

Example

Solving a two-step equation

Solve $5 x - 7 = 18$ .

Add 7 to both sides to remove the constant term from the left:
$5 x - 7 + 7 = 18 + 7$
This simplifies to $5 x = 25$ .
Divide both sides by 5 to make the coefficient of $x$ equal to 1:
$5x5=255\frac{5x}{5}=\frac{25}{5}$
Therefore, $x = 5$ .
Check the solution in the original equation:
$5 (5) - 7 = 25 - 7 = 18$
The left-hand side equals the right-hand side, so the solution is correct.

A coefficient is a number multiplying a variable. In $5 x$ , the coefficient of $x$ is 5.

Common Mistake

Changing signs without a reason

Collecting unknown terms

When the unknown occurs on both sides, collect all terms involving the unknown on one side and all constant terms on the other.

You may choose either side for the unknown. It is often convenient to choose the side that gives a positive coefficient.

Example

Collecting unknowns on one side

Solve $7 x - 4 = 3 x + 20$ .

Subtract $3 x$ from both sides to collect the unknown terms on the left:
$7 x - 3 x - 4 = 20$
This gives $4 x - 4 = 20$ .
Add 4 to both sides:
$4 x = 24$
Divide both sides by 4:
$x = 6$
Check using both sides of the original equation:
$7 (6) - 4 = 38$
and
$3 (6) + 20 = 38.$
Both sides have the same value, confirming the solution.

Tip

Keep the coefficient positive

If possible, collect the unknown terms on the side with the larger coefficient. This often avoids working with a negative coefficient, although either method is valid.

Equations containing brackets

Before collecting terms, remove brackets by expanding them. Expanding means multiplying every term inside a bracket by the factor outside it.

For example:

3 (x - 4) = 3 x - 12

If there is a minus sign before a bracket, every sign inside the bracket changes when it is expanded:

- (2 x - 5) = - 2 x + 5

Expanding before solving

Example

Solving an equation with brackets

Solve $4 (2 x - 3) = 3 (x + 6)$ .

Expand both brackets:
$8 x - 12 = 3 x + 18$
Subtract $3 x$ from both sides:
$5 x - 12 = 18$
Add 12 to both sides:
$5 x = 30$
Divide both sides by 5:
$x = 6$
Substitute $x = 6$ into the original equation. The left-hand side is $4 (12 - 3) = 36$ , and the right-hand side is $3 (6 + 6) = 36$ , so the solution is correct.

Common Mistake

Incomplete expansion

In $4 (2 x - 3)$ , the 4 multiplies both terms, giving $8 x - 12$ . Writing $8 x - 3$ leaves the second term unmultiplied.

Equations involving fractions

A linear equation may contain fractional coefficients or algebraic fractions. You can often simplify the working by multiplying every term by a common multiple of the denominators.

The lowest common multiple, or LCM, is the smallest positive number divisible by every denominator.

Example

Clearing fractional coefficients

Solve

x−13+x+22=6.\frac{x-1}{3}+\frac{x+2}{2}=6.

The LCM of the denominators 3 and 2 is 6. Multiply every term on both sides by 6:
$6(x−13)+6(x+22)=6(6).6\left(\frac{x-1}{3}\right)+6\left(\frac{x+2}{2}\right)=6(6).$
Cancel the denominators and simplify:
$2 (x - 1) + 3 (x + 2) = 36.$
Expand and collect like terms:
$2x−2+3x+6=365x+4=36.\begin{aligned} 2x-2+3x+6&=36\\ 5x+4&=36. \end{aligned}$
Subtract 4 and then divide by 5:
$\Rightarrow x=\frac{32}{5}.$
Substitution into the original equation gives
$325−13+325+22=95+215=6,\frac{\frac{32}{5}-1}{3}+\frac{\frac{32}{5}+2}{2} =\frac{9}{5}+\frac{21}{5} =6,$
so the exact solution is $x=325x=\frac{32}{5}$ .

Common Mistake

Multiplying only part of an equation

When clearing fractions, multiply every term on both sides by the LCM. Brackets help to show that an entire numerator, such as $x - 1$ , is being treated as one expression.

Special outcomes

Most linear equations have exactly one solution, but simplifying may sometimes eliminate the unknown completely.

No solution

If simplification produces a false statement, such as $0 = 5$ , no value of the unknown can satisfy the original equation.

Infinitely many solutions

Example

Recognising a special outcome

Consider $3 (x + 2) = 3 x + 8$ .

Expand the left-hand side:
$3 x + 6 = 3 x + 8.$
Subtract $3 x$ from both sides:
$6 = 8.$
This statement is false, so the equation has no solution. The two sides always differ by 2, regardless of the value of $x$ .

Common Mistake

Do not force a numerical answer

If all the unknown terms cancel, interpret the remaining statement. A false statement means no solution; a statement that is always true means infinitely many solutions.

Forming an equation from a context

In a modelling problem, the equation may not be given to you. You must translate the information into algebra.

A good model should:

define the unknown clearly;
express the relevant quantities in terms of that unknown;
use the relationship stated in the problem;
produce an answer that is reasonable in context.

Example

Modelling a ticket purchase

Adult tickets cost £8 and student tickets cost £5. A group buys 14 tickets for £88. Find the number of student tickets.

Let $x$ be the number of student tickets. Since there are 14 tickets altogether, the number of adult tickets is $14 - x$ .
Form an equation for the total cost:
$5 x + 8 (14 - x) = 88.$
Expand and simplify:
$5x+112−8x=88112−3x=88.\begin{aligned} 5x+112-8x&=88\\ 112-3x&=88. \end{aligned}$
Subtract 112 from both sides, then divide by negative 3:
$\Rightarrow x=8.$
The group therefore buys 8 student tickets and 6 adult tickets. Checking the cost gives £40 plus £48, which is £88.

Key Idea

Interpret the solution

A reliable solving routine

For a complicated linear equation, use this order:

Expand any brackets.
Clear fractions if this makes the equation simpler.
Collect like terms on each side.
Collect unknown terms on one side.
Collect constants on the other side.
Divide by the coefficient of the unknown.
Check by substitution into the original equation.

Exam technique

In the exam

Show balanced algebraic steps clearly; unexplained jumps make sign and arithmetic errors harder to detect.
Keep answers exact unless a decimal is requested, so write $325\frac{32}{5}$ rather than an unnecessarily rounded value.
Substitute your result into the original equation, especially when the equation contains negatives, brackets or fractions.
In a modelling question, define the unknown, form the equation and finish with a sentence interpreting the solution in context.

Self review

Check yourself

Can you solve $5 (2 x - 1) - 3 = 4 x + 16$ and verify your result?
What conclusion should you reach if an equation simplifies to $0 = - 7$ ?
How would you remove the denominators from $x4−x−36=2\frac{x}{4}-\frac{x-3}{6}=2$ ?

What you'll learn

What is a linear equation?

Equality and balancing

Solving a two-step equation

Collecting unknown terms

Equations containing brackets

Expanding before solving

Equations involving fractions

Special outcomes

No solution

Infinitely many solutions

Forming an equation from a context

A reliable solving routine

Review 1.2.2 Solve linear equations by teaching Genie

1.2.2 Solve linear equations

What you'll learn

What is a linear equation?

Equality and balancing

Solving a two-step equation

Collecting unknown terms

Equations containing brackets

Expanding before solving

Equations involving fractions

Special outcomes

No solution

Infinitely many solutions

Forming an equation from a context

A reliable solving routine

Review 1.2.2 Solve linear equations by teaching Genie

1.2.2 Solve linear equations

What you'll learn

What is a linear equation?

Equality and balancing

Solving a two-step equation

Collecting unknown terms

Equations containing brackets

Expanding before solving

Equations involving fractions

Special outcomes

No solution

Infinitely many solutions

Forming an equation from a context

A reliable solving routine

Review 1.2.2 Solve linear equations by teaching Genie

1.2.2 Solve linear equations Revision Guide

1.2.2 Solve linear equations

What you'll learn

What is a linear equation?

Equality and balancing

Solving a two-step equation

Collecting unknown terms

Equations containing brackets

Expanding before solving

Equations involving fractions

Special outcomes

No solution

Infinitely many solutions

Forming an equation from a context

A reliable solving routine

Review 1.2.2 Solve linear equations by teaching Genie

1.2.2 Solve linear equations Revision Guide