Linear Inequalities: Additional Practice (Set 1)

With $a = 0$ the inequality is $0 \ge b$, and here $b = 0$.

$0 \ge 0 \text{ is true} \;\Rightarrow\; \text{all real numbers}$

Unlike the strict $0 > 0$, which is false and gives no solution, the non-strict $0 \ge 0$ is true, so every real number is a solution.

Divide both sides by the positive number $b$, keeping the direction.

$bx > b^2 \;\Rightarrow\; x > \frac{b^2}{b} = b$

So the solution is $x > b$.

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

$x > 0:\; 5 \ge x \;\Rightarrow\; 0 < x \le 5 \qquad x < 0:\; 5 \le x \;\Rightarrow\; \text{no negative } x$

The positive case gives $0 < x \le 5$; the negative case asks for $x \ge 5$, impossible for $x < 0$. The solution set is $(0, 5]$.

The intersection keeps only what both sets contain.

$(-\infty, 2] \cap [2, \infty) = \{2\}$

The only number that is both at most $2$ and at least $2$ is $2$ itself, so the intersection is the single point $x = 2$.

The union sweeps together everything in either set.

$(-\infty, 2) \cup (2, \infty) = \{x : x \ne 2\}$

Every number except $2$ lies in one of the two open rays, and $2$ is in neither, so the union is all real numbers except $2$.

A leftover with no variable is judged outright.

$0 > -3 \text{ is true}$

Because the statement is true, the original held for every $x$, so the solution set is all real numbers.

The leftover has no variable, so its truth decides everything.

$0 > 3 \text{ is false}$

Because the statement is false, no value of $x$ works, so the solution set is empty.

The single step assumed $x > 0$. Redo the negative case with the flip.

$x < 0:\; 3 > 2x \;\Rightarrow\; x < \tfrac{3}{2}, \text{ true for all } x < 0$

Every negative number satisfies the original, so the honest answer is $x < 0$ or $x > \frac{3}{2}$. The naive step lost the whole negative ray.

Check each sign of $a$. A nonzero $a$ makes $ax$ take both signs as $x$ varies, so $ax \le 0$ cannot hold for all $x$.

$a = 0 \;\Rightarrow\; 0 \le 0 \text{ for every } x$

Only $a = 0$ makes the left side $0$ for all $x$, which always satisfies $\le 0$.

The multiplier $x$ has no fixed sign, so a single step cannot serve every $x$.

$\text{original } (-\infty, 3) \;\to\; \text{new } (0, 3)$

Writing $x^2 < 3x$ for all $x$ silently used the $x > 0$ branch, which is invalid for $x \le 0$, so those solutions were dropped.

A quotient is positive when its two parts share a sign.

$\text{both positive}: x > 1 \qquad \text{both negative}: x < 0$

Numerator $x$ and denominator $x - 1$ are both positive when $x > 1$ and both negative when $x < 0$; between $0$ and $1$ they differ in sign. So the solution is $x < 0$ or $x > 1$.

A quotient is negative when its two parts have opposite signs.

$x > 0 \text{ and } x + 2 < 0 \text{ is impossible}; \quad x < 0 \text{ and } x + 2 > 0 \Rightarrow -2 < x < 0$

The only region where numerator and denominator disagree in sign is $-2 < x < 0$, so that is the solution set.