How Do I Combine Two Inequalities Like 2 < x and x < 5?

You solved your inequality and got two conditions: 2 < x and x < 5. Now what? Do you add them? Multiply them? Write them on separate lines forever? The textbook answer uses a weird compressed notation — 2 < x < 5 — and leaves you wondering whether that is a real piece of algebra or a shorthand that only teachers use.

It is real algebra. And the reason it works is worth seeing clearly, because the same idea — two conditions on the same variable combine by taking their overlap — is the engine behind every compound inequality, every interval intersection, and most of the answers in the back of your algebra book.

The short answer

If 2 < x AND x < 5 must both be true at the same time, you write:

That is not a new kind of object. It is shorthand for "the two inequalities hold simultaneously," read left to right: 2 is less than x, and x is less than 5. As an interval, the solution is (2, 5) — every real number strictly between 2 and 5, with both endpoints excluded because both original inequalities are strict.

That's it. The rest of this article is why it works, when it doesn't work, and what the answer looks like when the two inequalities conflict.

Why AND means intersection

When you say "x satisfies both 2 < x and x < 5," you are asking: which numbers live in both solution sets at once?

• 2 < x picks out the ray (2, \infty).
• x < 5 picks out the ray (-\infty, 5).

A number has to be in both rays. The overlap — the intersection — is everything greater than 2 AND less than 5. Geometrically, shade both regions on the same number line and keep only what is shaded twice.

The chained form 2 < x < 5 is not a separate piece of notation with new rules — it is a typographic convenience that puts both conditions on one line. You read it as "x is between 2 and 5, exclusive," which is the same as "2 < x AND x < 5."

The three cases that cover everything

Whenever you combine two inequalities on the same variable with AND, one of three things happens. Which one depends only on whether the two rays point towards each other, away from each other, or the same way.

Case 1 — Same direction: the stronger one wins

Suppose you get x > 3 AND x > 7. Both conditions require x to be to the right of something. Which is harder to satisfy? Being greater than 7 is harder — any number that clears 7 automatically clears 3 as well. So the combination is:

The answer is (7, \infty). The weaker condition (x > 3) is absorbed by the stronger one. Geometrically: the ray (7, \infty) is inside the ray (3, \infty), so their overlap is the smaller one.

Same idea with <: if you have x < 10 AND x < 4, the answer is x < 4, i.e. (-\infty, 4). The smaller upper bound wins.

Case 2 — Opposite directions that meet: a bounded interval

This is the 2 < x < 5 case from the top of the article, now named. One inequality says "x must be above a," the other says "x must be below b," with a < b. The rays point towards each other and overlap in the middle:

The bracket shape of the answer follows the strictness of the input. If the inequalities were \ge and \le, the answer would be [3, 7] (closed). Mixed, like x \ge 3 AND x < 7, gives the half-open [3, 7).

Case 3 — Opposite directions that don't meet: no solution

What if the two conditions ask for the impossible? Like:

You want a number that is both less than 3 and greater than 7. There is no such number on the real line. The ray (-\infty, 3) and the ray (7, \infty) don't overlap — there is a gap between them from 3 to 7 where neither ray reaches.

The intersection is the empty set, written \varnothing or \{\}. In interval language: no interval at all. The system has no solution.

Do not try to write this as "7 < x < 3" — that notation is reserved for the case where the middle value genuinely lies between the two bounds. Writing 7 < x < 3 is a contradiction on its face (it would say 7 < 3, which is false), and the chained-inequality convention only makes sense when the left bound is actually less than the right bound. The correct answer is simply "no solution" or \varnothing.

The golden rule for chained inequalities

A chained inequality like a < x < b is only well-formed when a < b. The left-to-right reading has to be internally consistent. If your two conditions force a situation where a \ge b, you have a no-solution problem, and chained notation is wrong — you should write the conditions separately and conclude \varnothing.

This rule is the one place where the compressed notation can mislead you. The expression 7 < x < 3 looks like a valid double inequality but is self-contradictory; some students write it out without noticing, and then try to "solve" it. Always check: does the left endpoint lie below the right endpoint? If yes, you have an interval. If no, you have the empty set.

Worked conversion: inequality pair to interval

The three examples above, side by side:

One pair, one combined form, one interval. The structure is always the same: find the overlap of the two rays, and read it off.

A quick exam-style example

Problem. Solve 2x - 1 > 5 AND 3x + 4 \le 19.

Solve each piece.

First inequality. 2x - 1 > 5 \implies 2x > 6 \implies x > 3.

Second inequality. 3x + 4 \le 19 \implies 3x \le 15 \implies x \le 5.

You now have x > 3 AND x \le 5. Opposite directions, and 3 < 5, so they meet in a bounded interval. Chain them:

As an interval: (3, 5]. Hollow dot at 3 (strict), filled dot at 5 (non-strict).

Related doubts and where to go next

If you want to see the same idea with a slider — drag two inequalities around and watch the overlap and gap form and disappear — try the visualizer:

• AND vs OR: Intersection and Union of Intervals Visualizer.

For the mechanics of the brackets in (3, 5] vs [3, 5] vs (3, 5):

• [Round Bracket "(" vs Square Bracket "" in Interval Notation.

Once you are comfortable chaining AND-inequalities, you already know how to read an interval as a pair of simultaneous conditions — which is most of what the phrase "solution set" will ever mean in the rest of your algebra course.