How to Solve Inequalities and Write the Answer in Interval Notation

Inequality symbols and what the answer looks like

An inequality compares two expressions with one of four symbols: $<$ (less than), $>$ (greater than), $\le$ (less than or equal to) and $\ge$ (greater than or equal to). Its solution is usually an infinite set of numbers, so the answer is written as an inequality, a number line graph or an interval.

• $x < 5$ is the interval $(-\infty, 5)$; the round bracket means 5 is not included, shown by an open circle on a number line.
• $x \le 3$ is $(-\infty, 3]$; the square bracket means 3 is included, shown by a filled circle.
• $-2 \le x < 4$ is $[-2, 4)$.
• Infinity always gets a round bracket, because it is not a number that can be included.

The one rule that differs from equations

You may add or subtract any number on both sides, and multiply or divide both sides by a positive number, exactly as with equations. But multiplying or dividing both sides by a negative number reverses the direction of the inequality.

To see why, start with the true statement $-2 < 3$ and multiply both sides by $-1$. You get $2$ and $-3$, and $2 > -3$. Multiplying by a negative number reflects the number line, which swaps which side is larger. The same applies if you swap the two sides of an inequality: $5 > x$ means the same as $x < 5$.

Linear inequalities

Testing one value on each side of the boundary takes ten seconds and catches a forgotten sign flip every time.

Compound inequalities: and, or

A double inequality such as $-3 \le 2x + 1 < 9$ means both conditions hold. Solve it by doing the same operation to all three parts at once.

The answer is $[-2, 4)$, the overlap of the two conditions. An or statement, such as $x < -1$ or $x \ge 3$, is the union of two pieces and is written $(-\infty, -1) \cup [3, \infty)$.

Absolute value inequalities

The absolute value $|A|$ is the distance of $A$ from zero. A distance less than $k$ means being close to zero, which gives one interval; a distance greater than $k$ means being far away, which gives two pieces. For $k > 0$:

If $k$ is negative, $|A| < k$ has no solution and $|A| > k$ is true for every real number, since an absolute value is never negative.

Quadratic inequalities with a sign chart

Do not try to take square roots across an inequality. Instead, move everything to one side, find where the expression equals zero, and test the sign between those points.

Graphically, $y = x^2 - x - 6$ is an upward parabola that is above the x-axis outside its roots, which matches the sign chart. The quadratic calculator finds the roots if they are not easy to factor.

Rational inequalities

For an inequality with the variable in a denominator, never multiply both sides by that denominator: you do not know whether it is positive or negative. Use a sign chart with the zeros of the numerator and of the denominator instead.

Inequalities in two variables

An inequality such as $y > 2x + 1$ describes a region of the plane. Draw the boundary line $y = 2x + 1$, dashed for a strict inequality and solid for $\le$ or $\ge$, then shade the side that satisfies it. Testing the origin is quickest: $0 > 1$ is false, so shade the side that does not contain $(0, 0)$. The graphing calculator shades inequalities like this directly.

Common mistakes

• Forgetting to reverse the sign after multiplying or dividing by a negative number.
• Multiplying by a variable expression whose sign is unknown.
• Writing $|x - 4| < 3$ as $x - 4 < 3$ only, which loses the lower bound.
• Using a square bracket next to infinity, or the wrong bracket at an endpoint.
• Including a value that makes a denominator zero.

The inequality calculator shows each step and the interval notation. For the equation-solving moves used throughout, see how to solve linear equations.

A word problem with an inequality

Inequalities appear whenever a situation has a limit, a minimum or a budget. Phrases such as at least, no more than, at most and exceeds signal which symbol to use: at least means $\ge$, at most and no more than mean $\le$, and exceeds or more than means $>$.

The context added a restriction that the algebra alone did not give. It is common for real problems to need $t \ge 0$ or a whole-number condition, so always ask what values make sense before writing the final interval.

Checking an interval answer

An inequality answer has three parts that can each go wrong: the boundary values, whether each boundary is included, and which side of the boundary is kept. Check them in that order. Substitute each boundary into the original inequality to see whether it gives equality, which tells you whether the endpoint belongs in the answer. Then test one value inside each interval of your answer and one value outside.

For the rational example, $x = 1$ gives $\frac{0}{3} = 0$, and $0 \le 0$ is true, so 1 is included. At $x = -2$ the expression is undefined, so $-2$ is excluded even though the inequality allows equality. These two checks confirm the brackets in $(-2, 1]$.

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