Linear Inequalities: Practice

Dividing by $a$ silently assumes $a$ is positive.

$a < 0:\; x < 1 \qquad a = 0:\; \text{no solution}$

For a negative $a$ the direction must reverse, and for $a = 0$ the inequality is $0 > 0$, false. Only the case $a > 0$ actually gives $x > 1$.

The solution is empty when the coefficient vanishes and the statement becomes false.

$a - 2 = 0 \;\Rightarrow\; a = 2, \text{ giving } 0 > 0$

At $a = 2$ the inequality reads $0 > 0$, which is false, so there is no solution. For any other $a$ the coefficient is nonzero and the solution is a ray.

Exclude $x = 0$ and split on the sign of $x$.

$x > 0:\; 1 \ge x \;\Rightarrow\; 0 < x \le 1 \qquad x < 0:\; 1 \le x \;\Rightarrow\; \text{no negative } x$

The positive case gives $0 < x \le 1$; the negative case asks for $x \ge 1$, impossible for $x < 0$. The solution set is $(0, 1]$.

Exclude $x = 0$ and split on the sign of $x$.

$x > 0:\; 1 > -x \;\Rightarrow\; \text{all } x > 0 \qquad x < 0:\; 1 < -x \;\Rightarrow\; x < -1$

For positive $x$ the inequality always holds; for negative $x$ the flip gives $x < -1$. The union is $x < -1$ or $x > 0$.

An unknown-sign multiplier fails two tests at once.

$x - 3 < 0 \text{ possible}, \qquad x - 3 = 0 \text{ possible}$

Where $x - 3$ is negative the symbol should flip, and where it is zero the inequality collapses. Since you cannot rule out either, the step is doubly barred; split into cases where the sign is known instead.

Solve each piece, then intersect.

$2x - 1 > 5 \Rightarrow x > 3 \qquad 3 - x > 0 \Rightarrow x < 3$

The 'and' needs $x > 3$ and $x < 3$ at once, and no number is both, so the intersection is empty and there is no solution.

For $x < 0$, multiplying by $x$ reverses the direction.

$x < 0:\; 1 > x \;\Rightarrow\; x < 1$

The condition $x < 1$ is met by every negative number, so the whole case $x < 0$ survives and contributes all of $(-\infty, 0)$.

The chain means $x > 4$ and $x < -1$ at once.

$(4, \infty) \cap (-\infty, -1) = \varnothing$

The two rays do not overlap, so the intersection is empty. It is a genuine empty 'and', not a typo to be fixed by swapping the ends.

A leftover with no variable is judged true or false outright.

$4 < 1 \text{ is false} \;\Rightarrow\; \varnothing$

Because the statement is false, no value of $x$ satisfies that piece, so it contributes the empty set, the same as a contradiction from the equation lesson.

A true leftover means that piece holds for every $x$.

$-2 < 6 \text{ is true} \;\Rightarrow\; \text{that piece is } \mathbb{R}$

Intersecting anything with all of $\mathbb{R}$ leaves it unchanged, so the compound's solution set is simply the other piece's solution set.

Divide both sides by $-1$ (or multiply by $-1$), which reverses the direction.

$-x > -5 \;\Rightarrow\; x < 5$

The reversal is forced because the divisor is negative, so the solution set is $(-\infty, 5)$.

Divide by $a$, tracking the sign. A negative $a$ flips $ax > 0$ into $x < 0$.

$a < 0:\; ax > 0 \;\Rightarrow\; x < 0$

A positive $a$ gives $x > 0$ and $a = 0$ gives no solution, so only a negative $a$ produces exactly the negative numbers.