Solve equations with brackets - KS3 Maths

Key points

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An equation with brackets must be expanded before it is solved.

An understanding of expanding brackets is essential when solving equations with brackets.
An is like a set of scales that are in balance. The on either side of the equals symbol (=) have the same value as each other.
The equation needs to stay in balance. To keep the equation balanced, the same thing needs to be done to both sides.
Some equations have brackets. these brackets makes the equation easier to solve.
Expanding brackets does not change the value of an . It is two ways of writing the same expression.

Solving equations with brackets

Some equations include a set of brackets. When solving equations with brackets the first step is to expand the brackets.

When multiplying in brackets, make sure that everything inside the bracket is multiplied by the term (or number) outside the bracket.
Expand the brackets to produce an equation.
the equation to isolate the \(x\) term.
Divide both sides by the coefficient of \(x\) to find the value of \(x\)

Example

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Solve the equation 3(𝒙 + 5) = 18
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Expand the brackets on the left hand side of the equation. 𝒙 multiplied by 3 is 3𝒙. +5 multiplied by 3 is +15. The 18 is outside the bracket so remains as 18. The equation is now 3𝒙 + 15 = 18
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Subtract 15 from both sides. The equation is simplified to 3𝒙 = 3. The 𝒙 term has now been isolated.
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𝒙 multiplied by 3 is 3. Divide both sides by the 3 (the coefficient of 𝒙). The equation shows that 𝒙 = 1
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It is important to check the answer. The original equation stated that 3(𝒙 + 5) = 18. Substitute the value of 1 in place of the 𝒙 term.
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The solution is 𝒙 = 1. When 𝒙 is 1 then the value inside the brackets is 6. Multiply the value outside the brackets (3) with the value inside the brackets, 3 x 6 = 18. This confirms that the solution of 𝒙 = 1 is correct.

Question

Solve the equation 3(2\(x\) + 4) = 36

The equation: three open bracket two x plus four close bracket equals thirty six.

Solving equations where the coefficient of 𝒙 is negative

When solving equations using brackets, the coefficient of \(x\) may be negative.

The first step is to expand the brackets, taking care where negatives are involved.
the equation so that the coefficient of the unknown term is positive.
Simplify the equation to isolate the \(x\) term.
Divide by the coefficient of \(x\) to find the solution.
Check your solution by substituting your answer into the equation.

Example

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Solve the equation 2(11 – 3𝒙) = 4
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Expand the brackets on the left-hand side of the equation. 11 multiplied by 2 is 22. -3𝒙 multiplied by 2 is -6𝒙. The 4 is outside the bracket so remains as 4. The equation now reads as 22 – 6𝒙 = 4
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The coefficient of 𝒙 is negative (-6). Adding 6𝒙 to both sides of the equation keeps it in balance but makes sure that the coefficient of the unknown term is positive. The equation now reads as 22 = 4 + 6𝒙
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Isolate the 𝒙 term by subtracting 4 from both sides. The equation now reads as 18 = 6𝒙
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𝒙 multiplied by 6 = 18. Divide both sides by the 6 (the coefficient of 𝒙). The equation shows that 𝒙 = 3
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It is important to check the answer. The original equation stated that 2(11 – 3𝒙) = 4. Substitute the value of 3 in place of the 𝒙 term.
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When 𝒙 = 3, the equation becomes 2(11 – 3 × 3) = 4. The calculation inside the brackets is now 11 – 9, so the value inside the brackets is 2. Multiply the value outside the brackets (2) with the value inside the brackets (2). 2 × 2 = 4. When 𝒙 has a value of 3, then both sides of the equation are equal. This confirms the solution that 𝒙 = 3

Question

Solve the equation 3(11 − 2\(x\)) = 15

The equation: three open bracket eleven subtract two x close bracket equals fifteen.

Practise solving equations with brackets

Quiz

Practise solving equations with brackets with this quiz. You may need a pen and paper to help you with your answers.

Real-life maths

A pharmacist might use an equation involving brackets to calculate the correct amount of medicine needed for a patient according to their age.

If a pharmacist needed to work out the correct dose for an eight-year-old child, when the adult dose is 200 mg, they can use ‘Young’s formula’ to help them:

Child’s dose = Adult’s dose × (\( \frac{age}{age + 12} \))