One ray says x > 2, the other says x < 6. The solution to the system is only the stretch they share — the band from 2 to 6.
To ride the deep-water slide at the pool, two things must be true at once: you have to be over 12, and you have to know how to swim. Either one alone won't get you up the ladder — the lifeguard checks both. Lots of real conditions stack up like this. A single inequality describes one demand; a system of inequalities ties several demands together and asks for the values that satisfy all of them at the same time.
By the end of this lesson you'll be able to read a system written with a big curly brace, solve each inequality on its own (just as in 12.3), and then find the overlap on the number line — the one stretch that every condition agrees on. We'll keep the same colors throughout: the solution set is green, a region with no solution is red, a boundary value is amber, and the number line itself is blue.
12.4.1 Several conditions, all at once
Think back to an ordinary inequality like x > 2. It draws a line in the sand: every number to the right of 2 passes, everything else fails. Now imagine a second rule arrives — x < 6. To satisfy both, a number has to clear the first hurdle and the second. That little word and is the whole story of this lesson.
When two or more inequalities must hold together, we collect them with a big curly brace and call the bundle a system of inequalities:
The brace is shorthand for "and." A value is a solution of the system only if it satisfies every line at once.
This is the same brace you met in Stage 10 when you solved systems of equations — but the meaning is even more natural here. With equations, you hunted for the rare point that landed on both lines. With inequalities, each condition already allows a whole stretch of the line, so the system simply asks: which numbers are inside every stretch?
Key idea
A system means AND, never "or." Its solution set is the collection of values that make all the inequalities true together — the part the conditions share.
Watch out
"And" is stricter than each piece alone. Adding a second condition can only shrink the set of survivors, never grow it. If you ever end up with more numbers than one inequality allowed by itself, something has gone wrong.
🎮 Try it AND vs OR — who survives?
A number rolls in. Inequality A wants x > 2; inequality B wants x < 6. Slide the number and watch which conditions it passes — and whether it survives the system (an "and").
x =4
12.4.2 Finding the shared part on the number line
Here's the reliable recipe, and it never lets you down: solve each inequality by itself, draw each answer as a ray on the number line, then keep only the overlap — the stretch that every ray covers. Let's run the system from the picture above.
Start with the first line, 2x − 1 > 3. Add 1 to both sides to get 2x > 4, then divide by the positive 2: x > 2. (No flip — we divided by a positive number, just like in 12.2.) The second line, x − 2 < 4, gives x < 6 after adding 2. So the system becomes:
Each inequality is solved on its own first — exactly the work of 12.3. Only then do we hunt for the overlap.
Now stack the two rays. The blue ray x > 2 sweeps everything to the right of 2; the amber ray x < 6 sweeps everything to the left of 6. The numbers caught by both rays sit between the two boundaries — and that green band is the answer:
Stack the rays, then read off the overlap. The solution is 2 < x < 6 — open dots at both ends, because neither 2 nor 6 is itself allowed.
Worked example
Solve { x + 1 ≥ 0, x − 3 < 0 }.
First line: x + 1 ≥ 0 → subtract 1 → x ≥ −1 (a filled dot, since "≥" includes −1). Second line: x − 3 < 0 → add 3 → x < 3 (an open dot at 3). The overlap runs from −1 up to 3: −1 ≤ x < 3 — closed on the left, open on the right.
{ x ≥ −1, x < 3 } overlaps in the band −1 ≤ x < 3. The left end is filled (−1 is in), the right end is open (3 is out).
Watch out
The two ends of an interval can have different dots. Here −1 is included (a "≥") but 3 is excluded (a strict "<"). Copy the dot from whichever inequality made that boundary — don't make both ends match out of habit.
🎮 Try it Overlap finder
Pick a system. The widget solves each line, stacks the two rays, and shades the green overlap below — or warns you in red if the rays miss each other entirely.
System
12.4.3 Four quick patterns for the overlap
Once both inequalities are solved, every two-line system falls into one of just four shapes. You don't have to memorize them as a chant — each one is obvious the moment you picture the rays — but it's handy to name them so you can spot the answer at a glance.
Pattern 1 — both "greater than." If both conditions point right, the stricter one wins: a number that clears the larger bound automatically clears the smaller one too. So { x > 2, x > 5 } collapses to just x > 5 — take the larger lower bound.
Pattern 2 — both "less than." Mirror image: if both point left, keep the smaller ceiling. { x < 2, x < 5 } becomes x < 2 — take the smaller upper bound.
Pattern 3 — greater than the small one AND less than the big one. The rays come from opposite ends and meet in the middle. { x > 2, x < 5 } gives the finite band 2 < x < 5 — a real interval with two ends.
Pattern 4 — greater than the big one AND less than the small one. Now the rays point away from each other and never touch. { x > 5, x < 2 } demands a number bigger than 5 yet smaller than 2 — impossible. The answer is no solution (the empty set).
The four shapes side by side. Three give a solution (green); the fourth gives nothing at all — the rays point apart and share no ground.
System
Picture
Solution
x > 2 and x > 5
both point right
x > 5 (larger bound)
x < 2 and x < 5
both point left
x < 2 (smaller bound)
x > 2 and x < 5
meet in the middle
2 < x < 5
x > 5 and x < 2
point apart
no solution
Key idea
For two "same-direction" rays, the stronger demand swallows the weaker one (larger lower bound, or smaller upper bound). For two rays facing inward you get a finite band; for two facing outward you get nothing. Always glance at the picture to confirm.
Watch out
"Both greater than → take the larger" sounds backward at first. But think it through: to beat both 2 and 5 you only need to beat 5, the higher hurdle. Clearing the higher hurdle clears the lower one for free.
🎮 Try it Four-pattern explorer
Tap one of the four shapes. The widget shows the rule, draws the two rays, and shades the resulting set — or flags the empty case in red.
Pattern
12.4.4 Pinning down a parameter's range
Sometimes a system carries an unknown letter — a parameter — and you're told something about its solutions. Maybe "this system does have solutions" or "it has none." Working backward from that fact pins the parameter into a range. The picture does all the heavy lifting.
Take the system { x > 1, x < m }, where m is a number we get to choose. One ray points right from 1; the other points left from m. They overlap only if the left-pointing ray reaches past 1 — that is, only if m sits to the right of 1:
Slide m. When m > 1 the rays overlap in a green band; when m ≤ 1 they pull apart and there's nothing left.
So the system { x > 1, x < m } has solutions if and only ifm > 1. Notice the boundary case: at exactly m = 1 the conditions become x > 1andx < 1 — no number is both at once — so even m = 1 gives the empty set. That's why the answer is the strict m > 1, not m ≥ 1.
Worked example
For what m does { x > 5, x < m } have no solution?
The rays overlap only when the ceiling m beats the floor 5, i.e. when m > 5. So there's no solution exactly when m ≤ 5. (For instance, { x > 5, x < 2 } with m = 2 is empty — that was Pattern 4.)
Watch out
The trickiest part is the boundary itself. Always test equality ("what if m = 1 exactly?") to decide whether the answer uses a strict > or an inclusive ≥. Here the boundary fails, so the bound stays strict.
🎮 Try it Parameter slider
Slide m across the line for the system { x > 1, x < m }. Watch the green overlap appear the instant m passes 1, and vanish into red the moment it drops to 1 or below.
m =4
★ The big ideas, in one breath
A system of inequalities is a stack of conditions joined by "and," and its solution is every value that satisfies all of them — the part their stretches share. Solve each inequality on its own (just as in 12.3), draw the rays, and keep only the overlap. Two same-direction rays collapse to the stronger one (larger lower bound or smaller upper bound); two rays facing inward give a finite band; two facing outward give no solution. And when a parameter hides in the system, slide it and watch the overlap appear or vanish — testing the boundary tells you whether the bound is strict or inclusive.
What's next
So far every stretch has come from a linear inequality — a single ray. Next, in 12.5, a single quadratic inequality can already split the line into two pieces. The overlap habit you just built will carry straight over: read the regions, then keep the ones that work.
✎ Exercises 12.4
In your own words: what does the curly brace in a system of inequalities mean — "and" or "or"? And what does that say about the size of the solution set as you add more conditions?
Show answer
It means and: a solution must satisfy every inequality at once. Because each new condition is another hurdle, adding conditions can only shrink the solution set, never grow it.
Solve the system { 2x − 1 > 3, x − 2 < 4 } and describe the solution on the number line.
Show answer
First line: 2x − 1 > 3 → 2x > 4 → x > 2. Second line: x − 2 < 4 → x < 6. Overlap: 2 < x < 6 — a band with open dots at 2 and 6.
Solve { x + 1 ≥ 0, x − 3 < 0 }. Be careful with the two endpoints.
Show answer
x + 1 ≥ 0 → x ≥ −1 (filled dot). x − 3 < 0 → x < 3 (open dot). Overlap: −1 ≤ x < 3 — closed on the left, open on the right.
Without solving anything new, give the solution of each "same-direction" system: (a){ x > 2, x > 5 }; (b){ x < 2, x < 5 }.
Show answer
(a) Both point right → take the larger lower bound: x > 5. (b) Both point left → take the smaller upper bound: x < 2.
Solve { x > 5, x < 2 }. Sketch the rays to explain your answer.
Show answer
One ray needs x bigger than 5; the other needs x smaller than 2. The right-pointing ray (from 5) and the left-pointing ray (to 2) point away from each other and never overlap. So there is no solution — the empty set.
Match each system to its pattern (both-greater / both-less / middle / none): (a) {x > 0, x < 4}; (b) {x < −1, x < 3}; (c) {x > 7, x < 1}; (d) {x > −2, x > 4}.
Show answer
(a) middle → 0 < x < 4. (b) both-less → x < −1. (c) none → no solution. (d) both-greater → x > 4.
Solve the three-line system { x > 0, x < 6, x > 2 }. (The overlap habit still works — just keep the part all three share.)
Show answer
x > 0 and x > 2 together need only x > 2 (the larger lower bound). Pair that with x < 6. Overlap of all three: 2 < x < 6.
For what values of m does the system { x > 1, x < m } have at least one solution?
Show answer
The rays overlap only when the ceiling m sits to the right of the floor 1, i.e. m > 1. At exactly m = 1 we'd need x > 1 and x < 1 at once — impossible — so m = 1 is excluded. Answer: m > 1 (strict).
For what values of m does the system { x > 5, x < m } have no solution?
Show answer
It overlaps when m > 5, so it is empty exactly when the ceiling fails to clear the floor: m ≤ 5 (and m = 5 itself is empty, since x > 5 and x < 5 can't both hold).
A challenge of words: a ride requires you to be at least 12 years old andunder 60. Write the system for the allowed age a, then state the solution as an interval.
Show answer
"At least 12" → a ≥ 12 (inclusive); "under 60" → a < 60 (strict). System: { a ≥ 12, a < 60 }. Overlap: 12 ≤ a < 60 — 12 is allowed, 60 is not.
🎯 Quick check
Six questions to lock it in. Tap the answer you think is right.
§ For teachers and parents
This lesson addresses CCSS A-REI.B.3 (solve linear inequalities in one variable and represent the solution set), A-CED.A.3 (represent constraints by a system of inequalities and interpret solutions as the values satisfying all conditions), and 7.EE.B.4b (solve and graph one-variable inequalities arising from real situations). The #1 misconception is taking the union instead of the intersection — treating a system as "or" rather than "and," so that students report every number satisfying some inequality rather than only those satisfying them all. A close cousin is mishandling the empty case (insisting on an interval when the rays don't overlap). The antidote is physical and visual: a system means all at once, so shade each ray on the same number line and keep only the part they share. If the arrows never cover common ground, there is no solution. Encourage learners to always test one endpoint and one interior number to confirm both the direction and whether each boundary is included.
eastmath.com · Stage 12 · 12.4 Systems of Inequalities · Intuition before notation
eastmath.com · 12.4 Systems of Inequalities · 12.4.2 Finding the shared part on the number line