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      Solving linear equations where brackets are used

      I can recognise that there is more than one way to remove a bracket when solving an equation.

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      Lesson 14 of 17
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      • Year 8

      Solving linear equations where brackets are used

      I can recognise that there is more than one way to remove a bracket when solving an equation.

      Lesson details

      Key learning points

      1. It is possible to divide by the coefficient of the bracket first when solving an equation.
      2. When there are common factors this may be the most efficient method.
      3. When the coefficient is a fraction it may be helpful to rewrite the equation.
      4. It is possible to expand the bracket before solving an equation.
      5. You will need to be able to make decisions about the most efficient method to a solution.

      Keywords

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

      Common misconception

      That all brackets have to be expanded before we solve an equation with brackets it in.

      How would you solve $$2y=50$$? You would divide both sides by $$2$$. So why not start with 'divide $$2$$' when solving $$2(y-7)=50$$?

      Use examples like $$2(y-7)=50$$ and get pupils to solve them first by expanding and then again by multiplying by the reciprocal. Get pupils to compare the efficiency of both methods.
      Teacher tip

      Licence

      This content is © Oak National Academy Limited (2025), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

      Lesson video

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      Prior knowledge starter quiz

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      6 Questions

      Q1.
      $$3\over2$$ is the of $$2\over3$$.

      opposite
      fraction
      Correct answer: reciprocal
      product

      Q2.
      Expand $$7(2x-3)$$.

      $$7x-21$$
      $$14x-3$$
      Correct answer: $$14x-21$$
      $$14x+21$$
      $$-14x-21$$

      Q3.
      Expand $${1\over5}(10x-25)$$.

      $$50x-125$$
      $$50x+125$$
      $$2x-25$$
      $$2x+25$$
      Correct answer: $$2x-5$$

      Q4.
      The solution to $$8y+20=4$$ is $$y=$$ .

      Correct Answer: -2, negative 2, negative two, - 2

      Q5.
      Solve $$7x-3=2x+12$$.

      Correct answer: $$x=3$$
      $$x=2$$
      $$x=-2$$
      $$x=-3$$

      Q6.
      Solve $$7x-3=2x+11$$.

      $$5\over14$$
      $$5\over8$$
      $$8\over5$$
      Correct answer: $$14\over5$$

      Assessment exit quiz

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      6 Questions

      Q1.
      $$8(x+5)=5(2x-4)$$ is an example of .

      an expression
      an unknown
      Correct answer: an equation
      a substitution

      Q2.
      Which of these show a correct method and solution for the equation $$9(y-3)=45$$?

      $$y-3=45$$, therefore $$y=48$$
      Correct answer: $$y-3=5$$, therefore $$y=8$$
      $$9y-3=45$$, therefore $$9y=48$$ and $$y={48\over9}$$
      Correct answer: $$9y-27=45$$, therefore $$9y=72$$ and $$y=8$$
      $$9y-27=45$$, therefore $$9y=18$$ and $$y=2$$

      Q3.
      The solution to $$8(x+5)=5(2x-4)$$ is $$x=$$ .

      Correct Answer: 30, thirty, x=30

      Q4.
      Solve $$7(x+5)=5(2x-4)$$.

      $$x=-{55\over3}$$
      $$x=-5$$
      $$x=5$$
      Correct answer: $$x={55\over3}$$

      Q5.
      Which is the most efficient start to solving the equation $${5\over4}(3x+1)=20$$?

      multiply out the brackets
      divide both sides by $$5$$
      multiply both sides by $$5\over4$$
      Correct answer: multiply both sides by $$4\over5$$
      multiply both sides by $$4$$

      Q6.
      The solution to the equation $${5\over4}(3x+1)=20$$ is $$x=$$ .

      Correct Answer: 5, five, x=5

      Lesson appears in

      UnitMaths / Solving linear equations

      Maths

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