How to Solve Inequalities Step by Step

An inequality compares values using signs such as less than, greater than, less than or equal to, and greater than or equal to. Suppose you need to solve x + 4 > 9. Subtract 4 from both sides to get x > 5. Every number greater than 5 satisfies the original statement.

You can use the same balancing idea as an equation. If 3x <= 18, divide both sides by 3 to get x <= 6. Adding, subtracting, multiplying, or dividing both sides by the same positive number preserves the comparison.

This is the main method for solving inequalities: simplify, perform inverse operations, and keep the two sides balanced. The answer is usually a range of possible values rather than one single number. You can often check the direction by asking whether a larger or smaller value should make the original statement true.

The special rule appears when you multiply or divide by a negative number. For example, solve -2x > 8. Dividing both sides by -2 gives x < -4. The greater-than sign reverses and becomes less than.

Why does the sign flip? Start with a true statement such as 3 > 1. Multiply both sides by -1 and you get -3 < -1. On a number line, multiplying by a negative reflects values across zero, which reverses their order.

This rule is the most common mistake in solving inequalities. The sign does not reverse because you moved a term across the equals area or because a negative number appeared somewhere in the question. It reverses only when both sides of the inequality are multiplied or divided by a negative quantity.

A number line makes an inequality answer visible. For x > 5, place an open circle at 5 because 5 itself is not included, then shade to the right. For x <= 6, use a filled circle at 6 because equality is allowed, then shade to the left.

You can check your algebra by choosing a value from the shaded region and substituting it into the original inequality. If x < -4, try x = -5 in -2x > 8. The left side becomes 10, and 10 > 8 is true. Then try a value outside the solution, such as 0. The statement becomes 0 > 8, which is false.

Compound inequalities can describe a range between two boundaries. For example, 2 < x <= 7 means x is greater than 2 and at most 7. You can show both boundaries on one number line and shade only the values that satisfy both conditions. If a graph is required, label the boundary clearly before shading. If interval notation is required, translate the same boundary rules carefully so the algebra and graph still describe the same set of values. When solving inequalities, write one algebra step per line. That makes it easier to spot the exact moment a negative division happens and prevents a silent sign error.

Inequalities are the natural tool whenever a situation involves a limit. Suppose a phone plan costs 12 pounds a month plus 50 pence for each gigabyte of extra data, and you want to spend no more than 20 pounds. Let x be the extra gigabytes. The condition is 12 + 0.5x <= 20. Subtract 12 to get 0.5x <= 8, then divide by 0.5 to find x <= 16. In context, x must also be zero or more, so a sensible answer is any amount from 0 up to 16 gigabytes.

One trap appears when the variable itself could be negative. Consider 1/x > 2. It is tempting to multiply both sides by x, but you do not know whether x is positive or negative, so you cannot tell whether the sign should flip. Reason it through instead. If x were negative, 1/x would be negative and could not exceed 2, so x must be positive. Multiplying by a positive x then gives 1 > 2x, so x < 1/2. The full solution is 0 < x < 1/2.

Inequalities with squared terms need extra care. Consider x squared < 9. A common mistake is to write x < 3 and stop, or to write x < plus or minus 3, which has no clear meaning. Test some values. If x is 2 or minus 2, x squared is 4, which works. If x is minus 4, x squared is 16, which does not. The solution is every number between minus 3 and 3, written -3 < x < 3. For x squared > 9, the solution splits into two separate pieces: x < -3 or x > 3.

For a general quadratic, factorise and think about the graph. To solve x squared - x - 6 > 0, factorise to (x - 3)(x + 2) > 0. The graph of y = (x - 3)(x + 2) is a U shape crossing the x axis at -2 and 3. It lies above the axis outside those points and below it between them. The answer is therefore x < -2 or x > 3. Check by substituting a value from each region, such as -3, 0 and 4, and confirming which ones make the original statement true.

Absolute value inequalities are easiest to solve if you read the absolute value as a distance. The expression |x - 3| means the distance between x and 3 on a number line, so |x - 3| < 5 asks for every number less than 5 units away from 3. Count 5 units each way from 3 and you reach -2 and 8, so the solution is -2 < x < 8. Algebraically, this is the same as writing -5 < x - 3 < 5 and then adding 3 to all three parts.

When the inequality points the other way, the solution splits into two separate pieces. Consider |2x + 1| >= 7. Either 2x + 1 >= 7, which gives x >= 3, or 2x + 1 <= -7, which gives x <= -4. The solution is x <= -4 or x >= 3: two separate rays on the number line, with filled circles at -4 and 3. Check one value from each region and one in between. When x = 0, |1| = 1, which is not at least 7, confirming that the middle section is excluded, while x = 5 gives 11, which works.

Inequalities with two variables describe whole regions of a graph rather than stretches of a number line. To show y > 2x + 1, first draw the boundary line y = 2x + 1. Use a dashed line, because points exactly on it do not satisfy a strict inequality, and a solid line for less than or equal to, or greater than or equal to. Then pick a test point off the line, usually (0, 0). Substituting gives 0 > 1, which is false, so the origin is not in the solution, and you shade the side of the line away from it, here the region above.

Several inequalities together mark out a feasible region where all of them hold, the basis of linear programming. Suppose a bakery makes x cakes and y pies, with x + y <= 10 because of oven space and 2x + y <= 14 because of staff time, while x >= 0 and y >= 0. If a cake earns 5 pounds profit and a pie 3, the best plan lies at a corner of the region. Checking the corners (0, 0), (7, 0), (4, 6) and (0, 10) gives profits of 0, 35, 38 and 30, so 4 cakes and 6 pies is best.

Solving inequalities is mostly about the same balanced operations used for equations. Isolate the variable, but reverse the inequality sign whenever you multiply or divide both sides by a negative number. Use a number line and a quick substitution check to confirm that your solution range points in the correct direction.