Algebra Foundations / Module 3, Equations / Lesson 1
Solving Linear Equations in One Variable
Learn to solve linear equations with a balance method you can trust: one-step, two-step, brackets, fractions and word problems, with worked examples, graded practice and full answers.
- Mathematics
- Secondary and high school, ages 13 to 15
- About 50 minutes
- Interactive solver, worked examples and practice
1Learning goal: what you will be able to do
An equation is a statement that two things are equal. By the end of this lesson you will be able to find the value of the unknown that makes the statement true, prove that your answer works, and turn a real problem into an equation before you solve it.
Learning outcomes
- FoundationExplain an equation as a balance, and solve one-step and two-step equations using inverse operations.
- CoreSolve equations with the unknown on both sides, with brackets and with fractions, and check each answer by substitution.
- CoreForm an equation from a word problem, solve it, and state the answer in context.
- ExtensionRecognise equations with no solution or with infinitely many solutions, and rearrange a formula to make a different letter the subject.
Where this lesson fits
This is the first lesson of Module 3 in Algebra Foundations. The other lessons shown are planned and are not yet published.
- Module 2, Expressions and substitutionPlanned
- Module 3, Lesson 1: Solving linear equationsThis lesson
- Module 3, Lesson 2: Linear inequalitiesOpen
- Module 3, Lesson 3: Straight-line graphs and equationsOpen
- Module 4, Simultaneous equationsPlanned
Where this topic appears in courses
Solving linear equations is taught in lower secondary school and revised in every later algebra course. It sits within UK Key Stage 3 and GCSE Mathematics, Cambridge Lower Secondary and IGCSE Mathematics, WAEC junior and senior secondary mathematics, IB Middle Years and Diploma mathematics, and US Common Core standards 8.EE.C.7 and HSA-REI.B.3. Depth varies: some courses stop at brackets and fractions, others add formulae and literal equations. This lesson supports, and does not replace, the specification you are following, so check it for the exact requirements.
2A familiar example: the taxi fare
You are going to meet a friend by taxi. The driver charges a fixed $3 the moment you sit down, then $2 for every kilometre. You have exactly $17. How far can you go?
Predict first: how many kilometres can you travel?
Take a moment to estimate before you open this. Most people reach the answer without any algebra.
7 km. Take away the $3 fixed charge, which leaves $14. Split $14 into $2 pieces and you get 7. Check: $3 + 7 × $2 = $17.
Look at what you did. You did not guess and test. You undid the fare in reverse order: first you removed the fixed $3, then you divided by the $2 rate. That reversing is exactly what solving an equation means.
Algebra gives that thought a short form. Let k be the number of kilometres. The cost is $3 plus $2 for each kilometre, and it must equal $17:
3 + 2k = 17
The rest of the lesson makes this method reliable for any equation, including ones too tricky to do in your head. First, try it yourself with the balance solver below.
Try it: the balance solver
An equation is a balance: whatever you do to one side you must also do to the other. Choose an equation, then pick operations until x stands alone. The last preset is a trap, so look closely.
3Clear explanation
Unknowns, expressions and equations
A letter such as x stands for a number we do not know yet. A piece of algebra without an equals sign, such as 3x + 5, is an expression. Put an equals sign in and it becomes an equation, a statement that can be true or false depending on the value of x: 3x + 5 = 20. A solution is a value of x that makes the statement true. An equation is linear when x appears only to the first power, with no x2, no 1/x and no products of unknowns.
The balance principle
Key idea. If two quantities are equal, they stay equal when you do the same thing to both. Add the same number to both sides, subtract it, multiply both by it, or divide both by it (never by zero), and the equation stays true with the same solution.
Every step in solving an equation uses this one idea. The aim is to finish with x alone on one side, because then the other side is the answer. To get there, use the inverse operation, the one that undoes what has been done to x.
| Done to x | Undo it by | Example |
|---|---|---|
| Add 5 | Subtract 5 | x + 5 = 12 → x = 7 |
| Subtract 5 | Add 5 | x − 5 = 12 → x = 17 |
| Multiply by 3 | Divide by 3 | 3x = 12 → x = 4 |
| Divide by 3 | Multiply by 3 | x/3 = 12 → x = 36 |
When several things have been done to x, undo them in reverse order. In 3x + 5, the unknown was first multiplied by 3 and then had 5 added. So subtract 5 first, then divide by 3. This is the order-of-operations rule run backwards, and it is exactly what you did in the taxi example.
Layout. Write one equation per line, keep the equals signs in a column, and write the operation beside each line. Your working is then easy to check, and in an exam it earns method marks even if you slip on the arithmetic.
The unknown on both sides
When x appears on both sides, first gather the x terms on one side. Subtract the smaller x term from both sides, which keeps the coefficient positive and avoids sign errors. Then gather the plain numbers on the other side, and divide.
7x − 4 = 3x + 12 → 4x − 4 = 12 → 4x = 16 → x = 4
Brackets
Expand brackets first, multiplying the number outside by every term inside: a(bx + c) = abx + ac. Take particular care with a negative number outside: −2(x − 3) = −2x + 6, because −2 × −3 is +6. After expanding, treat the equation as one you already know how to solve.
Fractions
To clear fractions, multiply every term on both sides by the lowest common denominator, the smallest number that every denominator divides into. For x/3 + x/4 = 7, the lowest common denominator is 12, so multiply each of the three terms by 12.
Trap. Multiplying only the fraction terms is the most common error here. If the equation has a plain number on either side, that number must be multiplied too.
Check by substitution
Always substitute your answer into the original equation, not into one of your later lines, and work out both sides separately. If they match, the answer is right, whatever happened in the working. Checking takes ten seconds and catches almost every slip.
From words to equations
Many problems arrive as sentences. Use four steps every time: (1) define the unknown, with its unit; (2) translate the sentence into an equation; (3) solve it; (4) check against the original words and answer in a sentence.
| Words | Symbols | Watch for |
|---|---|---|
| 5 more than x | x + 5 | |
| twice x | 2x | |
| 5 less than x | x − 5 | The order is reversed from the words, so 5 − x is wrong. |
| is, gives, equals, makes | = | Marks where the equation splits into two sides. |
| the sum of three consecutive integers | n + (n + 1) + (n + 2) | Name the smallest, then build the others from it. |
Special cases and formulae Extension
Sometimes the x terms cancel completely. If what is left is false, such as 6 = 5, the equation has no solution. If what is left is always true, such as 3 = 3, the equation holds for every x and has infinitely many solutions. A formula is an equation with several letters, and the same steps let you make any letter the subject. You treat the other letters exactly as you would treat numbers.
Common mistakes
| Mistake | Why it is wrong | Correct idea |
|---|---|---|
| Changing only one side: 3x + 2 = 11 becomes 3x = 11 + 2 or 3x + 2 − 2 = 11. | The balance is broken, so the new equation has a different solution. | Subtract 2 from both sides: 3x = 9. |
| “Move it across and change the sign” without knowing why. | As a memory shortcut it hides the real step and invites sign slips. | Say what you do to both sides: “subtract 2 from both sides.” |
| Expanding one term only: 3(x + 4) = 3x + 4. | The 3 multiplies everything inside the bracket. | 3(x + 4) = 3x + 12. |
| Dividing one term only: 6x + 9 = 15 becomes 2x + 9 = 5. | Dividing both sides by 3 divides every term. | 2x + 3 = 5. |
| Clearing fractions on the fraction terms only. | Terms left unmultiplied no longer balance. | Multiply every term on both sides. |
| Subtracting a bracket badly: 5 − (x − 2) = 5 − x − 2. | The minus sign acts on both terms inside. | 5 − x + 2. |
| Stopping at −x = 4 and writing x = 4. | −x means −1 times x. | Divide both sides by −1: x = −4. |
| Never checking the answer. | Slips go unnoticed. | Substitute into the original equation. |
4Worked examples
Read each example with a pencil in hand. Cover the steps, try the first line yourself, then compare. Every example ends with a check.
Example 1 Foundation 2 marks: a two-step equation
Solve 3x + 5 = 20.
- The unknown was multiplied by 3, then 5 was added. Undo in reverse order, starting with the 5. Subtract 5 from both sides: 3x = 15.
- Divide both sides by 3: x = 5.
- Check in the original: 3(5) + 5 = 20, which matches the right side.
Answer: x = 5.
Example 2 Core 3 marks: the unknown on both sides
Solve 7x − 4 = 3x + 12.
- The smaller x term is 3x. Subtract 3x from both sides: 4x − 4 = 12.
- Add 4 to both sides: 4x = 16.
- Divide both sides by 4: x = 4.
- Check: left side 7(4) − 4 = 24, right side 3(4) + 12 = 24.
Answer: x = 4.
Example 3 Core 4 marks: brackets on both sides
Solve 3(2x − 5) = 2(x + 3) + 7.
- Expand the left bracket: 6x − 15. Expand the right bracket: 2x + 6, then add 7 to get 2x + 13. The equation is 6x − 15 = 2x + 13.
- Subtract 2x from both sides: 4x − 15 = 13.
- Add 15 to both sides: 4x = 28.
- Divide by 4: x = 7.
- Check: left side 3(14 − 5) = 3(9) = 27, right side 2(10) + 7 = 27.
Answer: x = 7.
Example 4 Core 4 marks: fractions
Solve x/3 + x/4 = 7.
- The lowest common denominator of 3 and 4 is 12. Multiply every term by 12: 4x + 3x = 84. (Here 12 × x/3 = 4x, 12 × x/4 = 3x and 12 × 7 = 84.)
- Combine like terms: 7x = 84.
- Divide by 7: x = 12.
- Check: 12/3 + 12/4 = 4 + 3 = 7.
Answer: x = 12.
Example 5 Core 5 marks: a word problem
The length of a rectangle is 3 cm more than twice its width. The perimeter is 54 cm. Find the width and the length.
- Define. Let w be the width in cm. Then the length is 2w + 3 cm.
- Translate. The perimeter is twice the sum of the two sides: 2(w + 2w + 3) = 54.
- Solve. Simplify inside the bracket: 2(3w + 3) = 54, so 6w + 6 = 54. Subtract 6: 6w = 48. Divide by 6: w = 8.
- The length is 2(8) + 3 = 19.
- Check. 2(8 + 19) = 54, and 19 is indeed 3 more than twice 8.
Answer: the width is 8 cm and the length is 19 cm.
Example 6 Extension 5 marks: special cases and a formula
(a) Solve 2(x + 3) = 2x + 5. (b) Solve 3(x + 1) = 3x + 3. (c) Make t the subject of v = u + at, then find t when v = 30, u = 6 and a = 8.
- (a) Expand: 2x + 6 = 2x + 5. Subtract 2x from both sides: 6 = 5. This is false for every x, so there is no solution.
- (b) Expand: 3x + 3 = 3x + 3. Subtract 3x from both sides: 3 = 3. This is true for every x, so there are infinitely many solutions.
- (c) Treat u and a like numbers. Subtract u from both sides: v − u = at. Divide both sides by a: t = (v − u)/a.
- Substitute: t = (30 − 6)/8 = 3. Check in the original formula: 6 + 8(3) = 30.
Answer: (a) no solution; (b) every value of x works; (c) t = (v − u)/a, and t = 3.
5Learner practice
Work each question on paper first, then enter your answer. You get two tries on each question. Two of the questions generate new numbers every time, so you can repeat them as often as you like.
Which is the best first step to solve 4x − 7 = 21?
Solve 5x + 8 = 43. Give the value of x.
Solve 9x − 5 = 4x + 20.
Solve 2(3x − 4) = 4x + 6.
A student solves 3(x + 4) = 18 like this: 3x + 4 = 18, so 3x = 14, so x = 14/3. Where is the first error?
Three consecutive whole numbers add up to 78. What is the smallest of the three?
How many solutions does 4(x − 2) = 4x − 8 have?
6Answer explanations
Attempt every question before you open these. A right answer with a wrong method is luck, so read each explanation even when you were correct. Questions 5 and 8 show their own worked solution as soon as you answer.
Question 1: the best first step for 4x − 7 = 21
Answer: add 7 to both sides. The 7 was subtracted from 4x, so the inverse operation is to add 7. It must be done to both sides, which gives 4x = 28.
Why the others fail: subtracting 7 would make the left side 4x − 14 and move away from the goal. Adding 7 to one side only breaks the balance. Multiplying by 7 is not the inverse of subtracting 7 and makes the numbers larger without isolating anything.
Question 2: 5x + 8 = 43
Subtract 8 from both sides: 5x = 35. Divide both sides by 5: x = 7.
Check: 5(7) + 8 = 43. A common slip is dividing the 5x by 5 but leaving the 8 untouched, which gives x + 8 = 43 and the wrong answer 35. Undo the addition before the multiplication.
Question 3: 9x − 5 = 4x + 20
Subtract 4x from both sides: 5x − 5 = 20. Add 5: 5x = 25. Divide by 5: x = 5.
Check: left side 9(5) − 5 = 40, right side 4(5) + 20 = 40. A common wrong answer is 25/13, which comes from adding the x terms instead of subtracting one from the other.
Question 4: 2(3x − 4) = 4x + 6
Expand: 6x − 8 = 4x + 6. Subtract 4x: 2x − 8 = 6. Add 8: 2x = 14. Divide by 2: x = 7.
Check: left side 2(21 − 4) = 34, right side 4(7) + 6 = 34. The usual slip is writing 6x − 4 for the left side, which means the −4 was not multiplied by 2.
Question 6: finding the error in 3(x + 4) = 18
Answer: the bracket was expanded incorrectly. The 3 multiplies both terms, so 3(x + 4) = 3x + 12, not 3x + 4. The later steps are all valid, so they do not contain the first error.
Correct solution: 3x + 12 = 18, so 3x = 6, so x = 2. Check: 3(2 + 4) = 18. The fraction 14/3 should itself be a warning, because a clean exam problem that produces an awkward fraction often hides a slip.
Question 7: three consecutive numbers that add to 78
Let n be the smallest. The three numbers are n, n + 1 and n + 2, and their sum is 78: 3n + 3 = 78. Subtract 3: 3n = 75. Divide by 3: n = 25.
The numbers are 25, 26 and 27, and 25 + 26 + 27 = 78. As a shortcut, the middle number is the mean, 78 ÷ 3 = 26, so the smallest is 25.
Question 9: 4(x − 2) = 4x − 8
Answer: infinitely many. Expand the left side: 4x − 8 = 4x − 8. The two sides are the same expression, so the statement is true for every x. Subtracting 4x leaves −8 = −8, which is always true.
Compare Example 6(a), where the leftover statement was false and there was no solution. An equation has exactly one solution only when the x terms do not cancel.
Questions 5 and 8: generated questions
The method is the same every time. For Question 5, subtract the smaller x term from both sides, move the plain number, then divide by the remaining coefficient. For Question 8, multiply every term by the lowest common denominator, combine like terms and divide.
Use New numbers to practise until you can do either without looking at the hint.
7Short recap
- An equation says two expressions are equal. A solution is a value of the unknown that makes it true.
- Do the same thing to both sides and the equation stays balanced.
- Undo operations in reverse order using inverse operations.
- With x on both sides, subtract the smaller x term first.
- Expand brackets completely before collecting terms.
- Clear fractions by multiplying every term on both sides by the lowest common denominator.
- Check by substituting into the original equation and evaluating each side separately.
- For words, define the unknown, translate, solve, check, and answer in a sentence.
- If the x terms cancel, a false leftover means no solution and a true leftover means infinitely many.
Tick what you can now do
If you ticked fewer than four, return to the matching worked example, redo it without looking, then try the generated questions again. Mastery here is the foundation for the next lessons in this module: linear inequalities and straight-line graphs.
Related Encyclopedia entries
- Algebra: the wider branch of mathematics this lesson belongs to.
- Linear Inequalities and Straight-Line Graphs and Equations: the next lessons in Algebra Foundations.
- Projectile Motion virtual lab: equations like these are used to model real motion.
References and further reading
Titles and publishers only. Search each site for the title to reach the exact page.
- OpenStax, Elementary Algebra 2e, the chapter on solving linear equations and inequalities. Free and open textbook. https://openstax.org/
- Khan Academy, Algebra 1, unit on solving equations and inequalities. Free video lessons and exercises. https://www.khanacademy.org/
- National Governors Association Center for Best Practices and Council of Chief State School Officers, Common Core State Standards for Mathematics, standards 8.EE.C.7 and HSA-REI.B.3. https://www.thecorestandards.org/
- UK Department for Education, Mathematics programmes of study: key stage 3. https://www.gov.uk/