Learn what an equation means and how to solve linear equations, equations with brackets, fractions and powers, quadratic and simultaneous equations, word problems and formulae, and how to spot no-solution cases and false solutions.
An algebraic equation is a statement that two expressions have the same value. Equations let us write down an unknown number, describe a relationship, and then find the value that makes the statement true. The real skill is not pushing symbols around. It is knowing why each step is allowed, keeping both sides equal at every step, and checking that the final answer works in the original equation. This guide starts with one-step equations and builds up to brackets, fractions, powers, quadratic equations, simultaneous equations, word problems, changing the subject of a formula, and equations that have no solution or an extra false solution. It has 28 worked examples checked by computer, three equation solvers, a practice drill, exam questions with answers, and a 20-question quiz.
An equation has two sides joined by an equals sign: left-hand side = right-hand side. The sign means that both sides have the same value. It does not mean “the answer comes next”. In 3x + 5 = 20, the letter x stands for an unknown number. To solve the equation is to find every value of x that makes it true.
Think of a pair of balanced scales. If you take 5 from one pan, you must take 5 from the other pan, or the scales tip. Every correct step in algebra does the same thing to both sides. Figure 1 shows this for 3x + 5 = 20.
| Term | Meaning | Example |
|---|---|---|
| Expression | A phrase with numbers and letters but no equals sign | 3x + 5 |
| Equation | Two expressions joined by an equals sign. True only for some values. | 3x + 5 = 20 |
| Identity | An equality that is true for every allowed value | 2(x + 3) = 2x + 6 |
| Formula | An equation that links real quantities | A = lw |
An expression is simplified or evaluated. An equation is solved. An identity needs no solving, because it is always true.
In 5a + 2b − 9 the terms are 5a, 2b and −9. The coefficient of a is 5, the coefficient of b is 2, and the constant is −9.
In 3x + 5x − 2, the terms 3x and 5x are like terms. They combine to give 8x − 2.
A solution is a value that makes the equation true. You can test any proposed solution by substituting it.
Left side: 2(4) + 3 = 8 + 3 = 11. Right side: 11. Both sides are equal, so x = 4 is a solution.
To find a solution, we undo the operations that were done to x. Operations that undo each other are called inverse operations.
| Operation | Inverse |
|---|---|
| Add a number | Subtract the same number |
| Subtract a number | Add the same number |
| Multiply by a number (not 0) | Divide by the same number |
| Divide by a number (not 0) | Multiply by the same number |
| Square | Square root (remember the ± sign) |
| Cube | Cube root |
(a) x + 7 = 15. Subtract 7 from both sides: x = 8.
(b) x − 9 = 4. Add 9 to both sides: x = 13.
(c) 6x = 42. Divide both sides by 6: x = 7.
(d) x/5 = 9. Multiply both sides by 5: x = 45.
In a two-step equation, work from the outside inwards. Undo the adding or subtracting first, then undo the multiplying or dividing. This is the reverse of the order in which the operations were applied.
(a) 3x + 4 = 19. Subtract 4: 3x = 15. Divide by 3: x = 5. Check: 3(5) + 4 = 19.
(b) 4x + 3 = −9. Subtract 3: 4x = −12. Divide by 4: x = −3. Check: 4(−3) + 3 = −12 + 3 = −9.
(c) 5 − 2x = 11. Subtract 5: −2x = 6. Divide by −2: x = −3. Check: 5 − 2(−3) = 5 + 6 = 11.
In part (c) the coefficient is negative, so the sign changes when you divide. Keep the sign with the term it belongs to.
When x appears on both sides, first collect the x terms on one side, then the numbers on the other. Doing the same operation to both sides is the reason that “change the side, change the sign” works.
(a) 5x + 2 = 2x + 17. Subtract 2x: 3x + 2 = 17. Subtract 2: 3x = 15. Divide by 3: x = 5. Check: 5(5) + 2 = 27 and 2(5) + 17 = 27.
(b) 7x − 4 = 3x + 12. Subtract 3x: 4x − 4 = 12. Add 4: 4x = 16. So x = 4. Check: 28 − 4 = 24 and 12 + 12 = 24.
(c) 4x + 1 = 10 − 2x. Add 2x: 6x + 1 = 10. Subtract 1: 6x = 9. So x = 1.5. Check: 4(1.5) + 1 = 7 and 10 − 3 = 7.
One sign went wrong. Which one?
The fix: the +2 on the left side must be subtracted from both sides, so it becomes −2 on the right side: 3x = 17 − 2 = 15, and x = 5. A number that moves across the equals sign changes sign.
To remove brackets, multiply every term inside by the number outside. This is the distributive law: a(b + c) = ab + ac.
(a) 3(x + 4) = 21. Expand: 3x + 12 = 21. Subtract 12: 3x = 9. So x = 3.
(b) 2(x + 3) + 4 = 18. Expand: 2x + 6 + 4 = 18, so 2x + 10 = 18. Then 2x = 8 and x = 4.
(c) 5(2x − 1) − 3(x + 2) = 24. Expand: 10x − 5 − 3x − 6 = 24. Collect: 7x − 11 = 24. Then 7x = 35 and x = 5. Check: 5(9) − 3(7) = 45 − 21 = 24.
A minus sign in front of a bracket changes the sign of every term inside it.
Simplify −(x − 5). Multiply each term by −1: −x + 5.
Solve 10 − (x − 3) = 4. Expand: 10 − x + 3 = 4, so 13 − x = 4. Then −x = −9 and x = 9. Check: 10 − (9 − 3) = 10 − 6 = 4.
What happened to the −3 inside the bracket?
The fix: the minus sign multiplies both terms. −(x − 3) = −x + 3, so 10 − (x − 3) = 10 − x + 3 = 13 − x. A good habit is to write the minus sign as “×(−1)” before you expand.
To remove fractions, multiply every term on both sides by a common multiple of the denominators. The lowest common multiple keeps the numbers small.
(a) x/4 + 3 = 8. Subtract 3: x/4 = 5. Multiply by 4: x = 20.
(b) x/3 + x/6 = 9. The lowest common denominator is 6. Multiply every term by 6: 2x + x = 54. So 3x = 54 and x = 18.
(c) (x + 3)/2 − (x − 1)/3 = 3. Multiply every term by 6: 3(x + 3) − 2(x − 1) = 18. Expand: 3x + 9 − 2x + 2 = 18, so x + 11 = 18 and x = 7. Check: 10/2 − 6/3 = 5 − 2 = 3.
(d) (x + 3)/4 = (x − 1)/2. Multiply by 4: x + 3 = 2(x − 1) = 2x − 2. So 5 = x. x = 5. Check: 8/4 = 2 and 4/2 = 2.
(a) 0.4x + 1.2 = 5.2. Subtract 1.2: 0.4x = 4. Divide by 0.4: x = 10.
(b) 1.5x − 0.75 = 3.75. Multiply every term by 100 to remove the decimals: 150x − 75 = 375. Then 150x = 450 and x = 3. Check: 4.5 − 0.75 = 3.75.
To undo a square, take the square root. A positive number has two square roots, so there are usually two solutions: x² = 49 gives x = 7 or x = −7, written x = ±7. A negative number has no real square root, so x² = −4 has no real solution. A cube is different: every real number has exactly one real cube root.
(a) x² = 49 gives x = ±7, because 7² = 49 and (−7)² = 49.
(b) x² = −4 has no real solution, because no real number squared is negative.
(c) (x − 3)² = 16. Then x − 3 = 4 or x − 3 = −4, so x = 7 or x = −1.
(d) x³ = 27 gives x = 3. (e) x³ = −8 gives x = −2.
(f) 2x² = 50. Divide by 2: x² = 25. So x = ±5.
Is 3 the only solution? Try x = 0.
The fix: x = 0 also works, because 0² = 3(0). Dividing by x lost that solution, and it is not allowed when x could be 0. Instead, move everything to one side and factorise: x² − 3x = 0, so x(x − 3) = 0, and x = 0 or x = 3.
A quadratic equation has the form ax² + bx + c = 0 with a ≠ 0. It can have two, one or no real solutions. There are three main methods: factorising, the quadratic formula, and completing the square. Whichever method you use, first write the equation with 0 on one side.
This rule works only when the product is zero. It is the reason a factorised quadratic gives separate solutions.
(a) x² − 5x + 6 = 0. Find two numbers with product 6 and sum −5: they are −2 and −3. So (x − 2)(x − 3) = 0 and x = 2 or x = 3.
(b) x² + 2x − 15 = 0. Two numbers with product −15 and sum 2: they are 5 and −3. So (x + 5)(x − 3) = 0 and x = −5 or x = 3.
(c) x² − 9 = 0. This is a difference of two squares: (x − 3)(x + 3) = 0, so x = ±3.
(d) x² = 3x. Then x² − 3x = 0, so x(x − 3) = 0 and x = 0 or x = 3.
(e) 2x² + 7x + 3 = 0. Split the middle term: 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3). So x = −1/2 or x = −3.
If the quadratic does not factorise easily, use the formula. It works for every quadratic.
Write a, b and c with their signs before you substitute. Put brackets round negative numbers.
(a) x² − 3x − 4 = 0: a = 1, b = −3, c = −4. b² − 4ac = 9 + 16 = 25. So x = (3 ± 5) ÷ 2, giving x = 4 or x = −1.
(b) 2x² + 5x − 3 = 0: a = 2, b = 5, c = −3. b² − 4ac = 25 + 24 = 49. So x = (−5 ± 7) ÷ 4, giving x = 1/2 or x = −3.
(c) x² − 4x + 1 = 0: a = 1, b = −4, c = 1. b² − 4ac = 16 − 4 = 12, and √12 = 2√3. So x = (4 ± 2√3) ÷ 2 = 2 ± √3. As decimals, x ≈ 3.732 or x ≈ 0.268 (3 decimal places).
Write the left side as a perfect square plus a number. Add the square of half the x-coefficient to both sides.
Solve x² + 6x + 1 = 0. Subtract 1: x² + 6x = −1. Half of 6 is 3, and 3² = 9. Add 9 to both sides: x² + 6x + 9 = 8, so (x + 3)² = 8. Then x + 3 = ±√8 = ±2√2. So x = −3 ± 2√2, which is x ≈ −0.172 or x ≈ −5.828 (3 decimal places).
The part under the square root, b² − 4ac, is the discriminant. Its sign tells you how many real solutions there are, before you solve anything.
| Discriminant | Real solutions | Graph |
|---|---|---|
| Positive | Two different | Crosses the x-axis twice |
| Zero | One repeated | Touches the x-axis |
| Negative | None | Does not reach the x-axis |
For what values of k does x² + kx + 9 = 0 have one repeated solution?
The discriminant must be zero: k² − 4(1)(9) = 0, so k² = 36 and k = 6 or k = −6. For example, k = 6 gives x² + 6x + 9 = (x + 3)² = 0.
Two equations with two unknowns are called simultaneous equations. A solution must make both equations true at the same time. On a graph, the solution is the point where the two lines cross (Figure 5).
Elimination adds or subtracts the equations so that one letter disappears. Substitution replaces one letter with an expression from the other equation.
(a) x + y = 11 and x − y = 3. Add the equations: 2x = 14, so x = 7. Then 7 + y = 11, so y = 4. Check the second equation: 7 − 4 = 3.
(b) 2x + 3y = 12 (1) and 3x − 2y = 5 (2). To make the y coefficients match, multiply (1) by 2 and (2) by 3: 4x + 6y = 24 and 9x − 6y = 15. Add them: 13x = 39, so x = 3. Put x = 3 in (1): 6 + 3y = 12, so y = 2. Check (2): 9 − 4 = 5.
Solve y = 2x + 1 and x + y = 10. Replace y in the second equation: x + 2x + 1 = 10. So 3x = 9 and x = 3. Then y = 2(3) + 1 = 7. This is the crossing point in Figure 5.
Solve y = x + 1 and y = x² − 5. Put the expressions equal: x + 1 = x² − 5. Rearrange: x² − x − 6 = 0, so (x − 3)(x + 2) = 0, giving x = 3 or x = −2. Use y = x + 1 to find y: when x = 3, y = 4; when x = −2, y = −1.
The solutions are (3, 4) and (−2, −1). Check (−2, −1) in the curve: (−2)² − 5 = −1. ✓
Two straight lines can also have no common point (parallel lines) or every point in common (the same line). In those cases the equations have no solution, or infinitely many solutions.
Algebra is most useful when it turns a real situation into an equation. Choose a letter for the unknown, write every other quantity in terms of it, form an equation from the statement, solve it, and then check the answer against the story.
| Words | Algebra |
|---|---|
| A number increased by 7 | x + 7 |
| A number decreased by 7 | x − 7 |
| Three times a number | 3x |
| A number divided by 4 | x/4 |
| Five less than twice a number | 2x − 5 (not 5 − 2x) |
| The sum of x and y | x + y |
| The difference between x and y | x − y or y − x, depending on the order stated |
| Two consecutive integers | x and x + 1 |
One number is 4 more than another and their sum is 30. Let the smaller number be x. Then x + (x + 4) = 30, so 2x = 26 and x = 13. The numbers are 13 and 17. Check: 13 + 17 = 30 and 17 − 13 = 4.
Ali is 3 times as old as his son. In 12 years, Ali will be twice as old as his son. Find their present ages.
Let the son’s age be x, so Ali’s age is 3x. In 12 years: 3x + 12 = 2(x + 12). Expand: 3x + 12 = 2x + 24. So x = 12. The son is 12 and Ali is 36. Check: in 12 years they are 24 and 48, and 48 = 2 × 24.
The length of a rectangle is 3 cm more than twice its width. Its perimeter is 54 cm. Let the width be w cm, so the length is (2w + 3) cm. Perimeter: 2(w + 2w + 3) = 54, so 3w + 3 = 27 and w = 8. The width is 8 cm and the length is 19 cm. Check: 2(8 + 19) = 54.
3 pens and 2 notebooks cost Rs 190. 2 pens and 5 notebooks cost Rs 310. Find the price of one pen and one notebook.
Let a pen cost Rs p and a notebook Rs n. Then 3p + 2n = 190 and 2p + 5n = 310. Multiply the first by 5 and the second by 2: 15p + 10n = 950 and 4p + 10n = 620. Subtract: 11p = 330, so p = 30. Then 3(30) + 2n = 190, so 2n = 100 and n = 50. A pen costs Rs 30 and a notebook Rs 50. Check the second statement: 2(30) + 5(50) = 60 + 250 = 310.
A taxi charges Rs 100 plus Rs 40 for each kilometre. The fare for a journey is Rs 420. How long is the journey? Let the distance be k km: 100 + 40k = 420. So 40k = 320 and k = 8. The journey is 8 km.
Many ratio and percentage questions are just equations in disguise. For a percentage, write the percentage as a decimal or fraction multiplier.
(a) 4 items cost Rs 12. Let 10 items cost Rs x. Then 4/12 = 10/x, so 4x = 120 and x = 30. (Check: each item costs Rs 3, and 10 × 3 = 30.)
(b) Which number has 20% equal to 18? 0.20x = 18, so x = 18 ÷ 0.20 = 90.
(c) After a 15% increase, a price is Rs 2300. What was the original price? The new price is 115% of the old one, so 1.15x = 2300 and x = 2300 ÷ 1.15 = Rs 2000. Check: 2000 × 1.15 = 2300.
A formula is an equation that links quantities. “Making w the subject” means rearranging so that w stands alone on one side. The balance rules are exactly the same as for solving equations: treat the wanted letter as the unknown and the other letters as numbers.
(a) A = lw. Divide by l: w = A/l (l ≠ 0).
(b) v = u + at. Subtract u: v − u = at. Divide by a: t = (v − u)/a (a ≠ 0).
(c) C = 5(F − 32)/9. Multiply by 9: 9C = 5(F − 32). Divide by 5: 9C/5 = F − 32. Add 32: F = 9C/5 + 32. Check: C = 100 gives F = 180 + 32 = 212.
(d) s = ut + ½at². Subtract ut: s − ut = ½at². Multiply by 2: 2(s − ut) = at². Divide by t²: a = 2(s − ut)/t² (t ≠ 0). Check with u = 2, t = 4, a = 3: s = 8 + 24 = 32 and 2(32 − 8)/16 = 3.
(a) 2x + 5 = 2x + 9. Subtract 2x from both sides: 5 = 9. This is never true, so there is no solution.
(b) 3(x + 2) = 3x + 6. Expand the left side: 3x + 6 = 3x + 6. This is true for every x, so there are infinitely many solutions. The equation is an identity.
A denominator can never be zero. Before you accept an answer, check that it does not make any denominator zero.
Solve 1/(x − 2) = 3. Here x cannot be 2. Take reciprocals: x − 2 = 1/3, so x = 7/3. This is allowed, because 7/3 − 2 = 1/3 is not zero. Check: 1 ÷ (1/3) = 3.
Some steps, such as squaring both sides, can create an extra value that does not satisfy the original equation. Such a value is called an extraneous solution. After squaring, always check each answer in the original equation.
Solve √(x + 2) = x. Square both sides: x + 2 = x². Rearrange: x² − x − 2 = 0, so (x − 2)(x + 1) = 0 and x = 2 or x = −1.
Check x = 2: √4 = 2. ✓. Check x = −1: √1 = 1, but the right side is −1. ✗.
The only solution is x = 2. The value −1 appeared because squaring cannot tell 1 from −1.
Every equation in two letters, such as y = 2x + 1, has a graph made of all the points that satisfy it. Solving two equations together means finding where the graphs meet. Solving a quadratic equation ax² + bx + c = 0 means finding where the curve y = ax² + bx + c meets the x-axis, as in Figure 3. So algebra and graphs are two views of the same idea. You can read more about drawing these graphs in Graphing Functions.
For thousands of years, equations were written in words. The word algebra comes from al-jabr, a term in the title of a book by the scholar Muhammad ibn Musa al-Khwarizmi, who worked in Baghdad in the early 9th century (around 820 CE). The word al-jabr is usually explained as “restoring”: moving a subtracted quantity to the other side of an equation. Indian mathematicians, including Brahmagupta in the 7th century, also gave rules for solving quadratic equations. The equals sign was introduced by the Welsh mathematician Robert Recorde in 1557. He chose two parallel lines of equal length because, he said, nothing could be more equal. The habit of using letters from the end of the alphabet, such as x, y and z, for unknowns is usually linked to the French mathematician René Descartes in the 17th century. Dates in early history are approximate.
Solve any equation of the form ax + b = cx + d. Type the four numbers (use 0 for a missing term). You will see the steps and a check. Try 2, 5, 2, 9 to see an equation with no solution.
Solve ax² + bx + c = 0. The tool shows the discriminant and says how many real solutions there are.
Solve two equations of the form ax + by = c. Type the six numbers (use 0 for a missing term).
Solve 4(2x − 3) + 5 = 3x + 20. Expand: 8x − 12 + 5 = 3x + 20, so 8x − 7 = 3x + 20. Subtract 3x: 5x − 7 = 20. Add 7: 5x = 27. Divide by 5: x = 27/5 = 5.4.
Check: left side = 4(10.8 − 3) + 5 = 4(7.8) + 5 = 36.2. Right side = 3(5.4) + 20 = 16.2 + 20 = 36.2. ✓
Why is this wrong?
The fix: the zero-product rule works only when the product is 0. Many pairs of numbers multiply to 6. Expand and move everything to one side first: x² − 5x + 6 = 6, so x² − 5x = 0, and x(x − 5) = 0. The solutions are x = 0 or x = 5. Check x = 5: (3)(2) = 6.
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1. Subtract 4x: 2x − 5 = 11. Add 5: 2x = 16. So x = 8. Check: 48 − 5 = 43 and 32 + 11 = 43.
2. Expand: 2x + 10 = 5x − 8. Subtract 2x: 10 = 3x − 8. Add 8: 18 = 3x. So x = 6. Check: 2(11) = 22 and 30 − 8 = 22.
3. Two numbers with product −12 and sum 1 are 4 and −3. So (x + 4)(x − 3) = 0 and x = −4 or x = 3.
1. Subtract 3x: 2x − 7 = 9. Add 7: 2x = 16. x = 8. Check: 40 − 7 = 33 and 24 + 9 = 33.
2. Expand: 6x − 3 − 2x − 8 = 9, so 4x − 11 = 9. Then 4x = 20 and x = 5. Check: 3(9) − 2(9) = 27 − 18 = 9.
3. Multiply every term by 6: 3(x + 3) − 2(x − 1) = 18. So 3x + 9 − 2x + 2 = 18 and x + 11 = 18. x = 7. Check: 5 − 2 = 3.
4. (a) (x − 5)(x + 2) = 0, so x = 5 or x = −2. (b) a = 2, b = 5, c = −3. b² − 4ac = 25 + 24 = 49. x = (−5 ± 7) ÷ 4, so x = 1/2 or x = −3.
5. From x − y = 2, y = x − 2. Substitute: 3x + 2(x − 2) = 16, so 5x − 4 = 16 and x = 4. Then y = 2. x = 4, y = 2. Check: 12 + 4 = 16.
6. Let the daughter’s age be x, so the mother is 4x. In 6 years: 4x + 6 = 3(x + 6) = 3x + 18. So x = 12. The daughter is 12 and the mother is 48. Check: in 6 years, 54 = 3 × 18.
7. (a) Multiply by 2: 2s = (u + v)t. Divide by (u + v): t = 2s/(u + v). (b) x − 3 = 1/2, so x = 3.5. Check: 1 ÷ 0.5 = 2.
The balance picture is the most useful idea for beginners. A learner who can say “I do the same to both sides” is less likely to make sign errors than one who memorises “change the side, change the sign”. Ask learners to predict a solution before they solve, and then to check it by substitution. For word problems, insist on three lines: “Let x be …”, the equation, and an answer sentence with units. The Linear Equation Solver, Quadratic Solver and Simultaneous Equations Solver on this page show working steps and a check, so learners can compare their own working with a worked solution. Encourage learners to find the mistake in the four Fix-It boxes before opening the answer.
This article is original EDUSAMBAM writing. Examination requirements differ between boards, so compare your practice with the specification of your own board.
20 questions covering equations, inverse operations, brackets, fractions, powers, quadratics, simultaneous equations, formulae and interpretation. Answer every question, then submit to see your score instantly.