An equation is used to show two expressions that are equal to each other.

Solving complex linear equations

Maths

Checking understanding of algebraic notation

Securing understanding of equality

Different types of equations

Equality in an equation

Solution to an equation

Many equations, one solution

Solving simple linear equations with an additive step

Solving simple linear equations with a multiplicative step

Preparing to solve two step linear equations

Solving two step linear equations

Rearranging to solve linear equations

Brackets in equations

Solving linear equations where brackets are used

Solving complex linear equations involving brackets

Solving linear equations graphically using technology

Problem solving with linear equations

In this unit, pupils learn how to rearrange an equation to find a specific value that satisfies the equation. In expressions and formulae, pupils use their understanding of rearrangement to change the subject of a formula.

In the prior unit, pupils saw that an arithmetic sequence can be represented graphically and that this is a linear relationship. Any point on the line satisfies this relationship. In this unit, pupils learn how to find a value that satisfies the relationship when the second value is known, without the aid of the graph.

In this unit pupils will learn how to find a value that satisfies a linear relationship when one value is known both with a single step operation, and multiple step operations. A variety of cases are explored including using brackets and rearranging.

Solving linear equations

$${{56}\over{y}}=7$$ is an example of a {{}} equation.

The solution to $${{6x−2}\over{4}}=7$$ is $$x=$${{}}

Which is the most efficient first step when solving the equation $${{x+7}\over{9}}={{x-1}\over{3}}$$ ?

Solve $${{x+7}\over{9}}={{x−1}\over{3}}$$. The solution is $$x=$${{}}

Which of these expressions are equivalent to $${{x+1}\over3}\times{2\over3}$$ ?

Solving complex linear equations

Lesson outcome