Linear Inequalities: Additional Practice (Set 2)

Treat $ax > a$ with coefficient $a$ and constant $a$. Split on the sign of $a$.

$a > 0:\; x > 1 \qquad a < 0:\; x < 1 \qquad a = 0:\; 0 > 0 \text{ is false}$

Dividing by a positive gives $x > \frac{a}{a} = 1$; a negative $a$ flips it to $x < 1$; and $a = 0$ gives the false $0 > 0$, so no solution. Cancelling $a$ would hide both the flip and the empty case.

With $a = 0$ the left side is $0$ for every $x$.

$0 \ge 3 \text{ is false}$

The statement is false regardless of $x$, so the solution set is empty.

All reals would need the coefficient to be zero while the statement is true. Here the coefficient and constant are the same expression $a - 2$.

$a = 2 \;\Rightarrow\; 0 > 0 \text{ is false}$

When the coefficient vanishes the constant vanishes too, giving the false $0 > 0$, so the strict inequality never yields every real number.

Divide both sides by the negative number $b$, which reverses the direction.

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

So the solution is $x < b$.

Start from the meaning of $a < b$ and take opposites.

$(-a) - (-b) = b - a > 0 \;\Rightarrow\; -b < -a$

Because $b - a$ is positive, the difference $(-a) - (-b)$ is positive too, so $-a$ is the larger and the order reverses. The flip is a forced consequence, not a decree.

Subtract $1$ and combine over the common denominator $x$.

$\frac{x + 1}{x} - 1 = \frac{1}{x} > 0 \;\Rightarrow\; x > 0$

The quotient $\frac{1}{x}$ is positive exactly when $x > 0$, so the solution is $x > 0$.

Solve each piece, then take the union.

$2x > 6 \Rightarrow x > 3 \qquad \text{or} \qquad x < 5$

Every real number is either greater than $3$ or less than $5$ (most numbers are both), so the union $(3, \infty) \cup (-\infty, 5)$ covers the whole line.

No solution requires the coefficient to vanish while the statement is false.

$k + 1 = 0 \;\Rightarrow\; k = -1, \text{ giving } 0 > 0$

At $k = -1$ the inequality reads $0 > 0$, which is false, so the solution set is empty.

Any $a \neq 0$ gives a ray, hence a solution. Only $a = 0$ can fail.

$a = 0 \;\Rightarrow\; 0 > 4 \text{ is false}$

With $a = 0$ and the positive right side $4$, the statement $0 > 4$ is false, so the solution set is empty. Only $a = 0$ does this.

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

$x > 0:\; 5 \le x \;\Rightarrow\; x \ge 5 \qquad x < 0:\; 5 \ge x \;\Rightarrow\; \text{all } x < 0$

The positive case gives $x \ge 5$ and the negative case gives every $x < 0$, so the union is $x < 0$ or $x \ge 5$.

Compare with the previous item: the only new thing is the sign of the right side.

$a = 0 \;\Rightarrow\; 0 > -4 \text{ is true}$

Here the $a = 0$ branch gives the true statement $0 > -4$, so it produces every real number, not an empty set. Any $a \neq 0$ gives a ray, so no value of $a$ leaves the inequality with no solution.

Work the negative case honestly, remembering the flip.

$x < 0:\; 2 < x \;\Rightarrow\; x > 2$

The requirement $x > 2$ cannot hold for a negative $x$, so the negative case gives nothing. Adding 'or $x < 0$' wrongly includes values that fail the original.