First-degree inequalities
Learning Objectives
- Solve a first-degree inequality and write its solution set as an interval.
- Apply the transformation rules, reversing the inequality when multiplying or dividing by a negative number.
- Read the solution of \(ax + b > 0\) off the sign table from Lesson 3.
- Explain why multiplying by a negative number reverses the inequality.
From "equals zero" to "greater than zero"
An equation \(ax + b = 0\) asks where a linear function is zero. An inequality asks where it is positive or negative: \(ax + b > 0\), or \(\geq, <, \leq\). We already built the tool for this in Lesson 3 — the sign of a linear function changes exactly once, at its zero \(x = -\tfrac{b}{a}\). Solving an inequality is just selecting the side of that zero where the sign is the one we want.
First-degree inequality
A first-degree inequality is one that can be written as one of
\[ax + b > 0, \qquad ax + b \geq 0, \qquad ax + b < 0, \qquad ax + b \leq 0,\]
with \(a \neq 0\). A solution is a real number that makes the statement true, and the solution set is the set of all of them. Unlike an equation, whose solution is usually a single point, the solution set of an inequality is typically a whole interval — a half-line.
Transforming an inequality
Two of the moves are just like those for equations; the third is new and is the one to watch.
Adding or subtracting the same quantity on both sides keeps the direction of the inequality.
Multiplying or dividing both sides by a positive number keeps the direction.
Multiplying or dividing both sides by a negative number reverses the direction (\(>\) becomes \(<\), \(\leq\) becomes \(\geq\), and so on).
A one-line check makes the third rule believable: start from the true statement \(2 < 5\) and multiply both sides by \(-1\). The numbers become \(-2\) and \(-5\), and indeed \(-2 > -5\). The order flipped.
Solving \(ax + b > 0\)
Let \(a \neq 0\) and write \(x_0 = -\tfrac{b}{a}\) for the zero. Then:
if \(a > 0\), the solution of \(ax + b > 0\) is \(x > x_0\), i.e. the interval \(\left(x_0, +\infty\right)\);
if \(a < 0\), the solution of \(ax + b > 0\) is \(x < x_0\), i.e. the interval \(\left(-\infty, x_0\right)\).
The other three inequalities follow the same pattern, with the endpoint included for \(\geq\) and \(\leq\) (a square bracket) and excluded for \(>\) and \(<\) (a round bracket).
Start from \(ax + b > 0\) and subtract \(b\): \(ax > -b\). Now divide by \(a\), and here the sign of \(a\) decides everything.
If \(a > 0\), dividing keeps the direction: \(x > -\tfrac{b}{a} = x_0\). If \(a < 0\), dividing by a negative number reverses it: \(x < -\tfrac{b}{a} = x_0\). This is exactly the sign table of Lesson 3 read as a statement about where \(f(x) > 0\): on the side of the zero carrying the sign of \(a\).
Three worked inequalities
Positive slope. \(2x - 6 > 0\). Add \(6\): \(2x > 6\); divide by \(2\) (positive, no flip): \(x > 3\). Solution set \(\left(3, +\infty\right)\). The sign table of \(2x - 6\) agrees: positive to the right of its zero \(3\).
The flip. \(-3x + 3 \geq 0\). Subtract \(3\): \(-3x \geq -3\); divide by \(-3\) — reverse — \(x \leq 1\). Solution set \(\left(-\infty, 1\right]\), with \(1\) included because of \(\geq\). Again the sign table agrees: with \(a < 0\), the function is positive to the left of its zero \(1\).
Reduce first. \(3x + 5 < 2x - 1\). Subtract \(2x\): \(x + 5 < -1\); subtract \(5\): \(x < -6\). Solution set \(\left(-\infty, -6\right)\).
Why does a negative multiplier flip the inequality?
This rule is the one learners forget, so it is worth seeing why it must hold rather than memorising it.
Multiplying by a number \(a\) is applying the function \(g(x) = ax\). In Lesson 3 we proved that this function is strictly decreasing when \(a < 0\). A decreasing function reverses order: if \(x_1 < x_2\), then \(g(x_1) > g(x_2)\). So multiplying an inequality through by a negative number must turn \(<\) into \(>\) — not as an arbitrary rule, but as a direct consequence of what "decreasing" means.
When \(a > 0\) the same function is increasing, which preserves order — that is why a positive multiplier leaves the inequality untouched. The two transformation rules are simply monotonicity, seen from the other side.
A note on the degenerate cases
If \(a = 0\), an inequality such as \(0 \cdot x + b > 0\) no longer depends on \(x\): it reduces to the plain statement \(b > 0\). If that is true, every real number is a solution (the solution set is \(\mathbb{R}\)); if it is false, there is no solution (\(\varnothing\)). As with equations, these all-or-nothing cases only occur when the inequality is not genuinely of the first degree.
Exercises
Exercise 1
Solve each inequality and choose its solution set.
Exercise 2
Solve each inequality; the direction is given, fill in the boundary value.
Exercise 3
Decide whether each statement about inequalities is true or false.
Exercise 4
Express each solution as an interval, minding open vs. closed endpoints.
Exercise 5
For each inequality, decide whether \(x = 5\) is a solution.
Summary
A first-degree inequality \(ax + b > 0\) (or \(\geq, <, \leq\)) is solved by isolating \(x\); its solution set is an interval (a half-line).
Adding to both sides and multiplying by a positive number keep the direction; multiplying or dividing by a negative number reverses it.
That reversal is just monotonicity: \(x \mapsto ax\) with \(a < 0\) is decreasing, and decreasing functions reverse order.
Endpoints are included for \(\geq, \leq\) (square bracket) and excluded for \(>, <\) (round bracket); \(\pm\infty\) always takes a round bracket. Equivalently, read the answer off the Lesson 3 sign table.