This Year 9 solving linear equations and rearranging formulae lesson is written for pupils aged 13-14 and follows the National Curriculum. In short, it shows you how to solve multi-step linear equations and change the subject of a formula. Before you start, it helps to revise percentage change and simple interest and expanding binomials and taking out common factors, because this lesson builds on them.
Year 9 solving linear equations and rearranging formulae lesson: what you will learn
This lesson covers solving linear equations (including brackets, unknowns on both sides and fractions), forming equations, and changing the subject of a formula.
Starter activity
A balance has two bags of marbles and three marbles on the left, and one bag and eight marbles on the right. The balance is level. Each bag holds the same number of marbles. How can you find how many are in a bag? Try taking the same thing off both sides.
Equations and the balance idea
Definition: equation, solving and inverse operations
An equation is a statement that two expressions are equal, such as \(3x + 5 = 20\). Solving means finding the value of the unknown that makes the statement true. Inverse operations (the opposite operation) undo what has been done to \(x\).
Rule: keep the balance
Think of a pair of balanced scales: whatever you do to one side you must do to the other, so that the scales stay level.
One- and two-step equations
Worked example 1: two steps
Solve \(3x + 5 = 20\).
Subtract 5 from both sides: \(3x = 15\). Divide both sides by 3: \(x = 5\). Check: \(3 \times 5 + 5 = 20\). Correct.
Worked example 2: a negative coefficient
Solve \(7 - 2x = 15\).
Subtract 7: \(-2x = 8\). Divide by \(-2\): \(x = -4\). Check: \(7 - 2 \times (-4) = 7 + 8 = 15\). Correct.
Equations with brackets
Method: equations with brackets
- Expand the brackets (or divide both sides by the number outside).
- Collect the \(x\) terms on one side and the numbers on the other.
- Solve and check in the original equation.
Worked example 3: a bracket
Solve \(3(x - 2) = 18\).
Expand: \(3x - 6 = 18\). Add 6: \(3x = 24\). Divide by 3: \(x = 8\). (A shortcut is to divide both sides by 3 first, giving \(x - 2 = 6\).)
Unknowns on both sides
Key fact: collecting terms
Collect the \(x\) terms on one side and the numbers on the other. It is neatest to move the smaller \(x\) term so that you keep a positive coefficient.
Worked example 4: both sides
Solve \(5x + 3 = 2x + 15\).
Subtract \(2x\): \(3x + 3 = 15\). Subtract 3: \(3x = 12\). So \(x = 4\). Check: left \(= 23\), right \(= 23\). Correct.
Worked example 5: brackets on both sides
Solve \(2(3x + 1) = 4(x + 5)\).
Expand: \(6x + 2 = 4x + 20\). Subtract \(4x\): \(2x + 2 = 20\). Subtract 2: \(2x = 18\). So \(x = 9\).
Equations with fractions
Rule: clearing fractions
To remove fractions, multiply every term on both sides by the lowest common multiple of the denominators.
Worked example 6: one fraction
Solve \(\dfrac{x}{4} + 3 = 8\).
Subtract 3: \(\dfrac{x}{4} = 5\). Multiply by 4: \(x = 20\).
Worked example 7: fractions on both sides
Solve \(\dfrac{2x - 1}{3} = \dfrac{x + 4}{2}\).
Multiply both sides by 6 (the LCM of 3 and 2): \(2(2x - 1) = 3(x + 4)\). Expand: \(4x - 2 = 3x + 12\). Subtract \(3x\): \(x - 2 = 12\). So \(x = 14\).
Forming and solving equations
Worked example 8: angles in a triangle
The angles of a triangle are \(x\), \(x + 10\) and \(2x - 10\) degrees. Find them.
Choose a letter for the unknown, write an equation, solve it and check. The angles add to 180°, so \(x + (x + 10) + (2x - 10) = 180\). Simplify: \(4x = 180\), so \(x = 45\). The angles are 45°, 55° and 80°, which add to 180°.
Formulae and changing the subject
Definition: formula and subject
A formula is a rule that links quantities, such as \(A = \tfrac12 bh\) for the area of a triangle, \(v = u + at\) for speed or \(F = \tfrac95 C + 32\) for converting temperatures. The subject of a formula is the letter on its own on one side.
Method: changing the subject
- Treat the formula like an equation.
- Use inverse operations, doing the same to both sides.
- Keep a whole side together over one fraction line.
- Check by substituting numbers.
Worked example 9: making \(x\) the subject
Make \(x\) the subject of \(y = 3x + 2\).
Subtract 2: \(y - 2 = 3x\). Divide by 3: \(x = \dfrac{y - 2}{3}\).
Worked example 10: making \(t\) the subject
Make \(t\) the subject of \(v = u + at\), then find \(t\) when \(v = 7\), \(u = 3\), \(a = 2\).
Subtract \(u\): \(v - u = at\). Divide by \(a\): \(t = \dfrac{v - u}{a}\). With the numbers: \(t = \dfrac{7 - 3}{2} = 2\).
Worked example 11: making \(h\) the subject
Make \(h\) the subject of \(A = \tfrac12 bh\).
Multiply by 2: \(2A = bh\). Divide by \(b\): \(h = \dfrac{2A}{b}\).
Watch out: common mistakes
- Doing an operation to only one side of the equation.
- Forgetting to multiply every term when clearing fractions.
- Dividing only the first term: from \(3x = 15 - 6\) you must divide the whole of the other side, not just one term.
- Sign errors when dividing by a negative number.
- When changing the subject, not dividing the whole side: \(t = \dfrac{v - u}{a}\), not \(v - \dfrac{u}{a}\).
Worked example 12: an equation with negative terms on both sides
Solve \(3-2x=11+2x\).
Add \(2x\) to both sides: \(3=11+4x\). Subtract 11: \(-8=4x\). Divide by 4: \(x=-2\). Check: left \(=3+4=7\), right \(=11-4=7\).
Worked example 13: a formula with a fraction
Make \(h\) the subject of \(A=\dfrac12(a+b)h\).
Multiply both sides by 2: \(2A=(a+b)h\). Divide by \((a+b)\): \(h=\dfrac{2A}{a+b}\). For \(A=30\), \(a=4\), \(b=6\): \(h=60\div10=6\).
Worked example 14: solving then substituting
Given \(y=3x+4\), find \(x\) when \(y=19\), and find \(y\) when \(x=-2\).
\(19=3x+4\), so \(3x=15\) and \(x=5\). When \(x=-2\): \(y=3(-2)+4=-2\).
Worked example 15: an equation from a word problem
A piece of ribbon 90 cm long is cut in two. One piece is 14 cm longer than the other. How long is each?
Let the short piece be \(x\) cm. \(x+(x+14)=90\), so \(2x=76\) and \(x=38\). The pieces are 38 cm and 52 cm.
Note: show every step
Write each line of working under the one before, and do the same operation to both sides, writing it at the side of the equation if it helps. This makes it easy to find an error. Always check by substituting your answer back into the original equation, not into a line you have already changed. If the two sides do not match, go back through your lines one at a time and look for a sign error, which is the most common mistake in equations. Finding your own mistakes is a key skill.
Watch out: negatives and fractions
- When you move a term to the other side its sign changes. Subtracting \(2x\) from both sides is not the same as dropping it.
- When you multiply to clear fractions, multiply every term, including whole numbers: \(\dfrac x2+3=5\) becomes \(x+6=10\).
- When you divide by a negative number, the answer changes sign: \(-3x=12\) gives \(x=-4\).
Did you know?
The equals sign = was first used in 1557 by a Welsh mathematician called Robert Recorde. He wrote that he chose two parallel lines of the same length "because no two things can be more equal". Before that, mathematicians wrote the word for equals in full, in Latin.
From words to equations
Method: forming an equation from a problem
- Read the whole problem and decide what you are trying to find. Give it a letter, such as \(x\), and write down what the letter stands for, including the units.
- Write every other quantity in terms of that letter. For example, "three more than" becomes \(x+3\) and "twice as many" becomes \(2x\).
- Look for a fact that gives two expressions with the same value, such as a total, a perimeter or an angle sum. Put an equals sign between them.
- Solve the equation step by step, doing the same to both sides.
- Answer the question that was asked, in words, and check the answer in the original problem, not just in your equation.
Worked example 16: consecutive numbers
Three consecutive whole numbers add up to 87. Find the numbers.
Let the smallest number be \(n\). The next two numbers are \(n+1\) and \(n+2\), because consecutive numbers go up by one each time. Their total is \(n+(n+1)+(n+2)=3n+3\), so the equation is \(3n+3=87\). Subtract 3 from both sides to get \(3n=84\), then divide by 3 to get \(n=28\). The numbers are 28, 29 and 30. Check: \(28+29+30=87\), so the answer is correct.
Worked example 17: a subject with a minus sign
Make \(x\) the subject of \(y=12-3x\).
The term containing \(x\) is negative, so it is easiest to add \(3x\) to both sides first: \(y+3x=12\). Next subtract \(y\) from both sides: \(3x=12-y\). Finally divide both sides by 3 to get \(x=\dfrac{12-y}{3}\). Check with a number: if \(x=2\), then \(y=12-6=6\), and the new formula gives \(x=\dfrac{12-6}{3}=2\), as it should.
Note: equation, formula, identity and expression
An expression such as \(3x+2\) has no equals sign. An equation such as \(3x+2=14\) is true only for particular values of the letter, which you find by solving it. A formula such as \(A=lw\) links two or more quantities and works for any values you substitute. An identity such as \(2(x+3)=2x+6\) is true for every value of \(x\). Using the right word helps you to understand what a question is asking you to do. "Solve" goes with an equation, "substitute" and "rearrange" go with a formula, "simplify" goes with an expression, and "show that" or "prove" usually goes with an identity. If you try to solve an identity, every number works, and if you try to solve an equation with no solution, the letters cancel and you are left with something false, such as \(3=5\).
Otto’s tip
Always check your solution by substituting it back into the original equation. If both sides match, you are done. Hoot hoot!
How to use this Year 9 solving linear equations and rearranging formulae lesson
First, read the key points so that you know where the Year 9 solving linear equations and rearranging formulae lesson is going. Then work through each definition and worked example, and cover the solution before you read it. After that, practise with the solving linear equations and rearranging formulae worksheet and check every question with the worksheet answers. When you feel ready, test yourself with the multiple choice questions or the quick quiz. Finally, sit the solving linear equations and rearranging formulae test and mark it with the test answers.
Also, the starters are a good warm-up, so you can keep the key facts fresh. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, this Year 9 solving linear equations and rearranging formulae lesson gives you everything you need for the chapter.
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