Linear Inequalities: Practice

Divide both sides by $-3$. Because the divisor is negative, reverse the symbol.

$\frac{-3x}{-3} < \frac{12}{-3} \;\Rightarrow\; x < -4$

The solution set is $(-\infty, -4)$. Keeping the symbol as $>$ would give the wrong half of the line.

Dividing by the negative number $a$ reverses the direction.

$ax > 2 \;\Rightarrow\; x < \frac{2}{a}$

The solution is the ray $\left(-\infty, \frac{2}{a}\right)$.

If $a \neq 0$ the solution is a ray, not all reals. If $a = 0$ the inequality reads $0 > 9$, which is false.

$a = 0 \;\Rightarrow\; 0 > 9 \text{ is false}$

Because the right side $9$ is positive, the $a = 0$ branch gives no solution, so no value of $a$ yields every real number.

With $a = 0$ the inequality reduces to a statement with no variable.

$a = 0 \;\Rightarrow\; 0 > -5 \text{ is true}$

Since $0 > -5$ holds, every real number is a solution. Any $a \neq 0$ gives only a ray, so $a = 0$ is the value that produces all of $\mathbb{R}$.

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

$x > 0:\; 3 > x \;\Rightarrow\; 0 < x < 3 \qquad x < 0:\; 3 < x \;\Rightarrow\; \text{no negative } x$

The positive case gives $0 < x < 3$; the negative case asks for $x > 3$, which no negative number meets, so it contributes nothing. The solution set is $(0, 3)$.

Check the original at $x = -2$.

$\frac{1}{-2} = -0.5 < 1 \text{ is true, yet } -2 > 1 \text{ is false}$

So $x = -2$ is a genuine solution that the step threw away. Multiplying by $x$ treated $x$ as positive and lost every negative solution.

An 'and' is the intersection of the two solution sets.

$(1, \infty) \cap (-\infty, 6) = (1, 6)$

The overlap of the two rays is the band $1 < x < 6$.

Dividing by $a$ is still the move, but a negative $a$ reverses the direction.

$ax > b \;\Rightarrow\; x < \frac{b}{a} \quad (a < 0)$

The boundary $\frac{b}{a}$ is the same, but the ray now runs to the left.

In the negative case, multiplying by $x$ reverses the symbol.

$x < 0:\; 4 < x \;\Rightarrow\; x > 4$

But $x > 4$ contradicts the assumption $x < 0$, so no negative value works and that case contributes nothing. All the solutions come from the case $x > 0$.

An 'or' is the union of the two rays.

$(3, \infty) \cup (5, \infty) = (3, \infty)$

Every number above $5$ is already above $3$, so the larger ray sits inside the smaller-bound one, and the union is simply $x > 3$.

An 'and' is the intersection of the two rays.

$(3, \infty) \cap (5, \infty) = (5, \infty)$

A number must beat both bounds, so the stricter one wins and the intersection is $x > 5$.

Check each sign of $a$. If $a > 0$ the solution is $x > 0$; if $a < 0$ it is $x < 0$; if $a = 0$ the inequality is $0 > 0$, which is false.

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

No case gives every real number, so no value of $a$ works.