Algebra 1
Compound inequalities
Real limits often come in pairs. A thermostat might need to keep a room between 60 and 80 degrees, two conditions at once. A ride might admit anyone older than 12 or with a guardian, where either condition will do.
Those two little words, and and or, are what this lesson is about, and they pull in opposite directions. And narrows things down to where both conditions hold, while or opens things up to where either one does.
Picture two bars you might have to clear. And is "you must clear both bars": the only numbers that survive are the ones that pass every condition, so you end up with the narrow strip where the conditions overlap. "Temperature between 60 and 80" is an and, since a reading has to be above 60 and below 80, so only the middle band counts.
Or is "clear either bar," so a number gets in if it passes even one condition, and you end up with everything either condition allows. "Water under 32 degrees or over 100" is an or: freezing or boiling, with the whole comfortable middle left out.
On a number line, you shade each condition and then read off the combination. For an and, you keep only the part where the two shadings overlap, typically a single segment between two endpoints 8.2.f1. For an or, you keep everything that either shading covered, often two separate rays heading opposite ways with a gap between them. The endpoints follow the same open/filled rule from Lesson 8.1: each boundary is open or filled according to its own sign.
In symbols, an and often gets written as one chained statement. "x > 2 and x ≤ 5," the numbers between 2 and 5 with 5 included, can be squeezed into 2 < x ≤ 5, read as a single sentence: x is bigger than 2 and at most 5.
A chained form like that always means and, and it's only well-formed when both signs point the same way. Something like 3 < x > 1 points two ways at once and doesn't say anything coherent, so don't write it. An or, by contrast, usually stays as two separate statements joined by the word "or," because its two pieces don't touch.
The solving follows the picture. For a chained and, you do the same move to all three parts at once, left, middle, and right together, because the whole statement has to stay balanced, not just the middle. For an or, you solve the two inequalities separately and then put their answers together.
New words
- 8.2.d1 Compound inequality: two inequalities combined into one statement.
- 8.2.d2 And (intersection): both parts must hold. 2 < x ≤ 5 is shorthand for x > 2 and x ≤ 5: the values between 2 and 5 (with 5 included). On the line, the overlap of the two shaded regions, a single segment.
- 8.2.d3 Double inequality (a.k.a. three-part inequality): the chained form a < x < b. It is only well-formed when both symbols point the same way, and it always means an and (a < x and x < b). Never write something like 3 < x > 1. A chained form with the symbols pointing opposite ways is meaningless.
- 8.2.d4 Or (union): at least one part holds. x < -1 or x > 3: everything in either ray; the two pieces don't touch.
Worked example
8.2.w1 Example 1, "and" (three-part). Solve -1 < x + 2 < 4. The middle has a +2 you want to clear, so subtract 2. To keep the whole chain balanced, subtract it from all three parts at once: -1 − 2 < x < 4 − 2, which is -3 < x < 2. Graph: open circles at -3 and 2 (both signs strict), shade the segment between them. Test a point from inside, x = 0: in the original, 0 + 2 = 2, and -1 < 2 < 4 is true. Solution: -3 < x < 2.
8.2.w2 Example 2, "and" with a division. Solve 3 ≤ 2x - 1 ≤ 9. Add 1 to all three parts: 4 ≤ 2x ≤ 10. Now divide all three by 2; it's a positive 2, so no sign flips: 2 ≤ x ≤ 5. Graph: filled circles at 2 and 5 (both inclusive), shade between. Solution: 2 ≤ x ≤ 5.
8.2.w3 Example 3, "or". Solve x < -1 or x > 3. This one is already solved. It's two separate conditions, so there's nothing to isolate. Graph: an open circle at -1 shading left, and an open circle at 3 shading right, with nothing in the middle. Solution: x < -1 or x > 3.
8.2.w4 Example 4, "or" needing work. Solve 2x ≤ -4 or x - 3 > 2. Solve each piece on its own. Left: divide 2x ≤ -4 by the positive 2 to get x ≤ -2. Right: add 3 to x − 3 > 2 to get x > 5. Then put them together. Graph: a filled circle at -2 shading left, and an open circle at 5 shading right. Solution: x ≤ -2 or x > 5.
The flip rule hasn't gone anywhere. It still applies to a chained inequality, and there it touches both signs at once.
8.2.w5 Example 5, three-part, divide by a negative (flip BOTH signs). Solve -6 < -2x ≤ 4. The middle is -2x, so divide all three parts by -2. Dividing by a negative flips each sign: the < and the ≤ both turn around, giving -6/-2 > x ≥ 4/-2, that is 3 > x ≥ -2. That's correct, but it's awkward to read with the bigger number on the left. Rewrite it left-to-right in increasing order: reading 3 > x ≥ -2 backwards gives -2 ≤ x < 3. Notice only the positions moved. The strict end stayed strict and the inclusive end stayed inclusive, so the 3 keeps its open mark and the -2 keeps its filled one. Graph: a filled circle at -2, an open circle at 3, shade the segment between. Now test, and check the endpoints too, since a flip is exactly where an endpoint can slip. Inside, x = 0: -6 < -2(0) = 0 ≤ 4, true. Endpoint x = -2: -2(-2) = 4, and -6 < 4 ≤ 4 is true, so -2 is in. Endpoint x = 3: -2(3) = -6, and -6 < -6 is false, so 3 is out. That matches the filled -2 and open 3. Solution: -2 ≤ x < 3.
Two mix-ups are worth naming now that you've seen the clean cases. The first is shading the whole line for an and: if you ever end up shading everything for an and, you've combined the pieces like an or by mistake, when an and keeps only the overlap.
The second shows up when the two conditions don't actually overlap. "x > 5 and x < 1" asks for numbers that are both above 5 and below 1, and there aren't any, so the answer is no solution, an empty graph rather than the whole line. (Swap the word to or and the same two pieces cover a great deal: everything below 1 together with everything above 5.)
Check yourself
- 8.2.c1 Rewrite 2 < x ≤ 5 as two separate inequalities joined by a word. Which word, and why is the graph a single segment instead of two rays? (It's x > 2 and x ≤ 5. The word is and, so only the numbers satisfying both survive, the overlap, which is the single segment from 2 to 5, open at 2 and filled at 5.)
- 8.2.c2 Why does x > 5 and x < 1 have no solution, while x > 5 or x < 1 has lots? Describe both graphs. (No number is above 5 and below 1 at once, so the and has no solution and its graph is empty. The or needs only one condition, so it covers everything below 1 together with everything above 5: two rays heading opposite ways.)
- 8.2.c3 You solve a three-part inequality and divide all parts by -2. What happens to the two inequality signs, and how would testing a point confirm it? (Dividing by a negative flips both signs at once. To confirm, pick a number from your final segment, put it in the original, and check it makes a true statement; testing the endpoints shows which is included and which isn't.)
These problems mix and with or on purpose, since having to tell them apart each time is what gets the difference to stick. The answers and the matching worked examples are at the end of the lesson if you get stuck. Notice as you go that and problems should collapse to one segment and or problems should open into two rays. If your picture comes out the other way, that's a cue to re-read the joining word.
(solve and describe the graph: segment-between for and, two rays for or)
Set A, "and"
Reveal answerHide to problem 1
-1 < x < 4 — add 1 to all parts; open circles at -1 and 4, shade between.Reveal answerHide to problem 2
-1 ≤ x ≤ 3 — subtract 1 from all parts; filled circles at -1 and 3, shade between.Reveal answerHide to problem 3
2 ≤ x ≤ 5 — add 1, divide by 2; filled circles at 2 and 5, shade between.Reveal answerHide to problem 4
-2 < x < 3 — divide all by 2; open circles at -2 and 3, shade between.Set B, "or"
Reveal answerHide to problem 5
x < -1 or x > 5 — two rays (open at -1 left, open at 5 right).Reveal answerHide to problem 6
x < -1 or x > 5 — left: x<-1; right: x>5; two rays.Reveal answerHide to problem 7
x ≤ -2 or x > 3 — filled at -2 left, open at 3 right.Reveal answerHide to problem 8
x ≥ 5 or x < -2 — filled at 5 right, open at -2 left.Set C, three-part, divide by a negative (flip BOTH signs, then rewrite in increasing order)