Systems of inequalities in one unknown
Learning Objectives
- State that the solution set of a system of inequalities is the intersection of the individual solution sets.
- Solve each first-degree inequality with a sign table and read off its interval.
- Intersect the intervals on the number line to solve the system.
- Solve a double inequality as a compact two-inequality system.
- Recognise when a system has no solution (incompatible).
Several conditions at once
A single first-degree inequality, solved in Unit 2, pins the unknown to an interval. But often a quantity must meet several requirements at the same time — "at least this, and at most that." Each requirement is an inequality, and together they form a system. We look for the values of \(x\) that satisfy all of them.
System of inequalities and its solution set
A system of inequalities in the unknown \(x\) is a collection of inequalities considered together. A solution of the system is a number that satisfies every inequality in it.
Since a solution must lie in the solution set of each inequality at once, the solution set of the system is their intersection:
\[S = S_1 \cap S_2 \cap \cdots\]
where \(S_1, S_2, \ldots\) are the solution sets of the individual inequalities. So the recipe is two steps: solve each inequality on its own, then intersect the intervals.
The tool for step one: the sign table
To solve each inequality we use the sign of its linear function, recorded in the sign table from Unit 1. For \(f(x) = ax + b\) with zero \(x_0 = -\tfrac{b}{a}\):
| \(x\) | \(-\infty\) | \(-\tfrac{b}{a}\) | \(+\infty\) |
|---|---|---|---|
| \(f(x)\) | opposite sign to \(a\) | \(0\) | same sign as \(a\) |
An inequality like \(f(x) > 0\) then selects the interval where the bottom row is positive; \(f(x) \geq 0\) adds the zero itself. (For a single linear inequality you may of course just isolate \(x\) instead — the two routes always agree.)
Two inequalities, one unknown
Solve the system \(\begin{cases} 3x - 6 > 0 \\ -2x + 8 \geq 0 \end{cases}\)
First inequality. Let \(f(x) = 3x - 6\); its zero is \(x = 2\), and \(a = 3 > 0\):
| \(x\) | \(-\infty\) | \(2\) | \(+\infty\) |
|---|---|---|---|
| \(f(x) = 3x - 6\) | \(-\,-\,-\) | \(0\) | \(+\,+\,+\) |
We need \(f(x) > 0\), so \(S_1 = (2, +\infty)\).
Second inequality. Let \(g(x) = -2x + 8\); its zero is \(x = 4\), and \(a = -2 < 0\):
| \(x\) | \(-\infty\) | \(4\) | \(+\infty\) |
|---|---|---|---|
| \(g(x) = -2x + 8\) | \(+\,+\,+\) | \(0\) | \(-\,-\,-\) |
We need \(g(x) \geq 0\), so \(S_2 = (-\infty, 4]\) (the positive part together with the zero).
Intersect. The system's solutions are the numbers in both sets:
\[S = S_1 \cap S_2 = (2, +\infty) \cap (-\infty, 4] = (2, 4].\]
On the number line, this is exactly where the two bars overlap — open at \(2\) (from the strict \(>\)) and closed at \(4\) (from the \(\geq\)):
Tidy up first, mind the flip
Solve \(\begin{cases} 2(x - 1) < x + 2 \\ 5 - x \leq 3 \end{cases}\)
Each inequality simplifies to first degree, so here we just isolate \(x\). First: \(2x - 2 < x + 2\) gives \(x < 4\), so \(S_1 = (-\infty, 4)\). Second: \(5 - x \leq 3\) gives \(-x \leq -2\); dividing by \(-1\) reverses it, \(x \geq 2\), so \(S_2 = [2, +\infty)\).
Intersecting, \(S = (-\infty, 4) \cap [2, +\infty) = [2, 4)\): closed at \(2\), open at \(4\). Each endpoint keeps the bracket it came in with.
A double inequality is a system in disguise
An expression like \(-1 \leq 2x - 3 < 5\) is read as two conditions at once: \(-1 \leq 2x - 3\) and \(2x - 3 < 5\). It is a system, written compactly.
Because the same operation can be applied to all three parts, we can solve it in one line — add \(3\) throughout, then divide by \(2\):
\[-1 \leq 2x - 3 < 5 \;\Longrightarrow\; 2 \leq 2x < 8 \;\Longrightarrow\; 1 \leq x < 4.\]
So \(S = [1, 4)\). (Were a step to require multiplying by a negative number, every part would flip — and the order of the two ends would reverse with them.)
When the conditions collide
Consider \(\begin{cases} x + 1 > 4 \\ 2x \leq 4 \end{cases}\). The first gives \(x > 3\); the second gives \(x \leq 2\). Intersecting,
\[S = (3, +\infty) \cap (-\infty, 2] = \varnothing.\]
No number is at once greater than \(3\) and at most \(2\): the system has no solution. Each inequality is perfectly solvable on its own — it is the demand to satisfy both at once that fails.
This is the same phenomenon we met with systems of equations in the previous lessons: a system can be incompatible. There, two parallel lines never met; here, two interval conditions face away from each other and share nothing. An empty intersection is a perfectly good answer — it says, precisely, that the requirements cannot all be met.
Exercises
Exercise 1
Solve each inequality, intersect, and choose the solution set.
Exercise 2
The system \(\begin{cases} 2x - 4 \geq 0 \\ -x + 6 > 0 \end{cases}\) has an interval solution. Find its endpoints.
Exercise 3
Decide whether each statement about systems of inequalities is true or false.
Exercise 4
Decide whether \(x = 3\) is a solution of each system (it must satisfy both inequalities).
Exercise 5
Solve \(1 < 3x - 2 \leq 7\) and give the endpoints of the solution interval.
Summary
A system of inequalities in one unknown asks for the values satisfying every inequality at once. Its solution set is the intersection \(S = S_1 \cap S_2 \cap \cdots\) of the individual solution sets.
Method: solve each inequality (a sign table, or by isolating \(x\)), then intersect the intervals on the number line. Each endpoint keeps the open/closed bracket it had.
A double inequality \(a \leq \text{expr} < b\) is a two-inequality system; it can be solved in one line by operating on all three parts (with the usual flip if you multiply by a negative).
The intersection may be empty: the system is then incompatible, with no solution — the inequality version of the incompatible systems from the previous lessons.