Equations and Inequalities

Updated 7 October 2026

An equation says that two expressions have the same value. Solving it means finding every value of the letter that makes that true. Every step has one justification: do the same thing to both sides, and the two sides stay equal. The one exception is an inequality, where dividing by a negative number reverses the sign.

How do negative numbers work in real situations?

A negative number is an amount on the other side of zero. A temperature of −4 °C is four degrees below freezing, a diver at −12 m is twelve meters below sea level, and a card worth −20 points takes twenty points off a score. On a number line the negative numbers sit to the left of zero, and every number is smaller than the numbers to its right. So −14 is colder than −11, because it lies further to the left, and the order from coldest to warmest of five freezers at 2 °C, −9 °C, −14 °C, −1 °C and −11 °C is −14, −11, −9, −1, 2. Below Zero introduces the line, and Ordering Negative Numbers puts it to work. Moving along the line is adding and subtracting, so a shed at −4 °C that warms by 9 °C reaches −4 + 9 = 5 °C.

A rise is an addition and a fall is a subtraction, wherever the number starts, which Adding and Subtracting Across Zero practices. If the shed then cools by 8 °C, it reaches 5 − 8 = −3 °C. Subtracting a negative number is the step that surprises people. If a player's score is 5 and a −20 card is taken away from them, the score goes up, because the card was costing them twenty points: 5 − (−20) = 5 + 20 = 25. Taking away a loss is a gain. Subtracting Negative Numbers works through this. The same rule finds the distance between two heights. From a cliff top 30 m above sea level down to a sea bed at −25 m is 30 − (−25) = 55 m. A distance is never negative, so it is the size of the difference, whichever way round the subtraction is done, which is what Absolute Value means.

Multiplying by a negative number is repeated change in the negative direction, as Multiplying and Dividing Negative Numbers shows. A quiz that gives −3 marks for each wrong answer takes 4 × (−3) = −12 marks for four wrong answers, and a pond losing 12 liters a minute changes by 5 × (−12) = −60 liters in five minutes. Two negatives multiply to a positive. Walking west is a negative speed and a time in the past is a negative time, so walking west at 4 meters a second, three seconds ago the walker was 12 meters east of where they are now: (−4) × (−3) = 12. Try it on the strip below by reversing the direction of walking, and then the clock.

These situations turn into equations with negative numbers in them, and the equations are solved in the usual way, as in Solving Linear Equations, with a letter such as r standing for the number that is not known (Algebraic Notation). On a quiz of 25 questions, with 3 marks for a right answer and −1 for a wrong one, a student who scores 51 and answers every question has r right and 25 − r wrong, so 3r − (25 − r) = 51. Expanding gives 4r − 25 = 51, so 4r = 76 and r = 19. Check it: 19 right answers earn 57 marks, 6 wrong answers lose 6, and 57 − 6 = 51. A pond of 180 liters that changes by −12 liters a minute is empty when 180 − 12m = 0, which is after m = 15 minutes. A time or a count that comes out negative, or not whole, is a sign that the equation was set up wrongly, so read the answer back into the situation.

How do you solve a linear equation?

Solve a linear equation by undoing the operations around the letter, one at a time, in reverse order. In 3x + 7 = 22, x was multiplied by 3 and then 7 was added, so subtract 7 from both sides first: 3x = 15. Then divide both sides by 3: x = 5. Doing the same to both sides keeps two equal amounts equal. A bracket changes the order of the work. In 4(x − 2) = 20, either expand to 4x − 8 = 20 or, since 4 divides 20, divide first to get x − 2 = 5, so x = 7. A negative coefficient needs care: in 10 − 2x = 4, subtract 10 to get −2x = −6, then divide by −2 to get x = 3. Finish by putting each answer back into its original equation: for x = 5, 3 × 5 + 7 = 22, and for x = 7, 4 × 5 = 20.

A bracket is expanded first, and the number in front multiplies every term inside it. See Equations with Brackets.

The same shortcut turns 3(x + 1) = 12 into x + 1 = 4. See Treating a Bracket as a Single Quantity.

When the letter appears on both sides, subtract the smaller x term from both sides so that it appears only once. Removing the smaller term keeps the coefficient positive. See Equations with the Unknown on Both Sides.

When the equation has fractions, multiply every term by the common denominator, including any term that is not a fraction. In $\frac{x}{3}+\frac{x}{4}=7$x/3 + x/4 = 7, multiply by 12: 4x + 3x = 84, so x = 12. See Equations with Fractions.

How many solutions does it have?

Usually the x terms survive and there is exactly one solution. Sometimes they cancel, and what is left decides. If it is false, such as 1 = 5, no value of x works. If it is true, such as 4 = 4, every value works, and the equation is an identity. See How Many Solutions an Equation Has.

In ax + 3 = 5x + 7, the x terms collect to (a − 5)x = 4. Choose a = 5 and the x term is zero, and zero lots of x can never equal 4. See Choosing a Coefficient to Fix the Solution Count.

How do you change the subject of a formula?

Changing the subject of a formula means rearranging it so that a different letter stands alone on one side. Undo the operations around the letter you want, in reverse order, doing the same to both sides. In v = u + at, subtract u and divide by a to get $t=\frac{\left(v-u\right)}{a}$t = (v − u)/a. To make r the subject of $A=\pi r^2$A = πr², divide both sides by $\pi$π, then take the square root: $r=\sqrt{\frac{A}{\pi }}$r = √(A/π). See Changing the Subject of a Formula. The steps change when the letter appears twice, as in ax = b + cx: gather both x terms on one side, take x out as a common factor, and divide by the bracket, so $x=\frac{b}{\left(a-c\right)}$x = b/(a − c). Numbers test a rearrangement. If u = 4, a = 2 and t = 3, then v is 4 + 2 × 3 = 10, and the new formula gives back (10 − 4) ÷ 2 = 3.

The same method handles any formula with a repeated letter. See Making a Twice-Appearing Letter the Subject.

Turning words into equations

A word problem becomes an equation once the unknown has a name. Write the definition down: let n be the smaller number. See Forming Equations.

A formula answers every version of a question at once, and a case you can count checks it. See Writing Formulas from Words.

How do you solve simultaneous equations?

To solve simultaneous equations, remove one unknown, solve for the other, then substitute back. Take x + y = 10 and x − y = 4. Adding the equations cancels y and leaves 2x = 14, so x = 7, and the first equation then gives y = 3. Adding is allowed because both equations hold at once, so their sides add to equal totals. On a graph each equation is a line, and the solution is where the lines cross. When no letter has matching coefficients, scale one equation first. With 2x + 3y = 12 and x + y = 5, multiply the second by 2 to get 2x + 2y = 10, then subtract: y = 2, and x = 3. Check both original equations: 2 × 3 + 3 × 2 = 12 and 3 + 2 = 5. For the first pair, 7 + 3 = 10 and 7 − 3 = 4.

Removing a letter by adding or subtracting the equations is called elimination. See Simultaneous by Elimination.

Sometimes both equations need scaling before a pair of coefficients match. See Solving Simultaneous Equations by Scaling.

Substitution rearranges one equation for one letter and puts that expression into the other. If y = x + 1 and 2x + y = 10, then 2x + x + 1 = 10, so x = 3 and y = 4. Use it when one equation already gives a letter on its own. See Simultaneous by Substitution.

A word problem with two unknowns needs two facts. Name both unknowns, write one equation from each fact, and read the answer back in the words of the question. See Writing a Pair of Equations from a Word Problem.

Two lines can cross once, run parallel, or be the same line: one solution, none, or infinitely many. Scaling one equation to match the other shows which. See How Many Solutions a Pair of Equations Has and Choosing a Coefficient in a Pair of Equations.

Sometimes the question asks only for x + y. If 3x + 2y = 11 and 2x + 3y = 9, adding gives 5x + 5y = 20, so x + y = 4. See Solving for x + y Without Finding x and y.

What does an inequality mean?

An inequality describes a range of values rather than one value. x > 4 means every number greater than 4. On a number line it is a ray, with a hollow circle at 4 because 4 is not greater than 4. $x\ge 4$x ≥ 4 includes 4, and the circle is filled. See Inequalities and Solution Sets of Inequalities. Solving one works like solving an equation: from 3x − 2 > 10, add 2 and divide by 3 to get x > 4. Adding to both sides, or multiplying by a positive number, keeps their order. Multiplying or dividing by a negative number reverses it, so the sign must flip: from −x < −4, dividing by −1 gives x > 4. Test one value on each side of the boundary. For x = 5, 3 × 5 − 2 = 13, which is more than 10; for x = 3, 3 × 3 − 2 = 7, which is not.

From 2x + 3 < 11, subtract 3, then divide by 2: x < 4. See Two-Step Inequalities.

Dividing by a negative number reverses the sign. From −2x > 6, divide by −2: x < −3. A negative sign reflects the number line: 2 < 3, but −2 > −3.

A compound inequality has two ends. $-3<x\le 5$−3 < x ≤ 5 has infinitely many solutions but only eight integer solutions, because −3 is out and 5 is in. See Integer Solutions of Inequalities.

Inequalities in two variables

A limit on a total of two things is written with an inequality sign: pens at $2 and pads at $5 with $30 to spend is $2p+5d\le 30$2p + 5d ≤ 30. "At most" gives $\le$≤ and "at least" gives $\ge$≥. See Forming an Inequality in Two Variables.

To graph it, draw the boundary line, dashed when the sign is strict and solid when it is $\le$≤ or $\ge$≥. Test a point off the line, usually the origin, and shade the side where the inequality holds. See Graphing an Inequality in Two Variables.

Two inequalities together are solved by the region where both shadings overlap. See The Overlap of Two Inequality Regions.

Equations with a modulus

The modulus |x| is the distance of x from zero, so it is never negative, and |x| = 5 has two solutions: 5 and −5. An equation with a modulus splits into two cases, because the expression inside the bars can be positive or negative. From |x − 3| = 5, either x − 3 = 5, giving x = 8, or x − 3 = −5, giving x = −2. Substitute both back to check. |x + 1| = −4 has no solution. See Solving Absolute Value Equations.

An inequality with a modulus is a statement about distance. |x − 3| < 5 means x is within 5 of 3, one stretch of the line: −2 < x < 8. |x − 3| > 5 means more than 5 away from 3, which is two separate rays: x < −2 or x > 8. See Solving Absolute Value Inequalities.

How do you know a solution is right?

A solution is right when it makes the original equation true, so substitute it into the equation you were given, not into a line you wrote along the way. Solving 5x + 4 = 2x + 19 gave x = 5. Work out each side separately: 5 × 5 + 4 = 29 and 2 × 5 + 19 = 29. The sides agree, so the answer stands. The first line matters because a slip in the middle carries into every later line, and those lines would agree with the wrong answer. For simultaneous equations, the pair must satisfy both equations, since a whole line of points fits either one alone. An inequality needs a different check, because its answer is a range: a value inside should work, and a value outside should fail. For x < −3 from −2x > 6, try x = −4, giving −2 × −4 = 8, which is more than 6, then try x = 0, which gives 0.

That test catches a sign that was not reversed.

The four mistakes worth naming

• Multiplying only some terms. Clearing fractions or expanding a bracket multiplies every term.
• Not reversing the sign. Dividing an inequality by a negative number reverses it every time.
• Leaving the letter undefined. "Let x be the number of adult tickets" is part of the answer.
• Not checking a modulus case. Substitute both cases back into the original equation.

Where this leads next

The algebra behind these methods is in algebraic expressions, indices and surds, and the arithmetic in arithmetic methods. Quadratic equations need factoring or the quadratic formula. Proportional relationships written as equations start in ratio and proportion.