Inequality Basics: Practice

The right end uses a strict $<$, so $2$ is excluded, while the left end includes $-3$.

$2 < 2 \;\text{ is false}$

The values $-3$, $0$, and $1$ all lie in the band, but $2$ does not.

'Above freezing' is strictly greater than $0$, so the left end is open. 'No higher than $30$' allows $30$, so the right end is inclusive.

$0 < t \le 30$

The left bound is strict and the right bound is inclusive.

Multiplying by $-1$ reflects both numbers across zero, which reverses their order.

$a < b \;\xrightarrow{\;\times(-1)\;}\; -a > -b$

The smaller number's opposite becomes the larger one.

Multiply both sides by $-\tfrac{3}{2}$ to isolate $x$. The multiplier is negative, so reverse $\ge$ to $\le$.

$-\tfrac{3}{2} \cdot \left(-\tfrac{2}{3}x\right) \le -\tfrac{3}{2} \cdot 4 \;\Rightarrow\; x \le -6$

The flip comes from multiplying by a negative.

The interval runs from the included $-5$ up to but not including $-2$.

$-5 \le -4 < -2 \;\checkmark$

The value $-2$ is excluded, $-6$ is below $-5$, and $0$ is above $-2$, so only $-4$ works.

Divide both sides by the positive number $2$, keeping the direction, then read from the other end.

$\frac{10}{2} > \frac{2x}{2} \;\Rightarrow\; 5 > x \;\Rightarrow\; x < 5$

Dividing by a positive number does not flip the symbol.

The least number is the one farthest to the left on the number line.

$-6 < -3 < -1$

So $-6$ is the smallest.

Combine the bounds into one line with the smaller bound on the left, keeping each inclusive symbol.

$x \ge 4 \text{ and } x \le 9 \;\Rightarrow\; 4 \le x \le 9$

Both ends are inclusive, so both use a closed symbol.

Substitute $x = -2$ into the left side and compare with the right.

$-2(-2) + 1 = 5, \qquad 5 > 4 \;\checkmark$

Since $5 > 4$ is true, $x = -2$ is a solution.

Both relations are 'greater', so both shade right from the same boundary $3$. The only change is at the boundary itself.

$x > 3 \text{: open circle} \qquad x \ge 3 \text{: filled circle}$

Switching strict to inclusive opens or fills the circle without moving the shading.

Both endpoints are inclusive, so both take brackets.

$-2 \le x \le 2 \quad\Longleftrightarrow\quad [-2, 2]$

A closed interval includes both of its endpoints.

The left condition $n > -3$ is strict, so $-3$ is excluded; the right condition $n \le 1$ is inclusive, so $1$ is included.

$-3 < n \le 1 \quad\Longleftrightarrow\quad (-3, 1]$

Open on the left, closed on the right.