Linear Inequalities: Practice

Adding a constant to both sides preserves the order of any two numbers, so it keeps the solution set, and it is undone by subtracting the constant.

$\text{add } 5 \;\Rightarrow\; \text{order preserved, solution set unchanged}$

Multiplying by $0$ collapses both sides and loses all information, and multiplying by $x$ or by $x - 1$ is illegal because those factors can be zero or negative, and a negative multiplier reverses the direction.

Scaling by a positive factor stretches or shrinks distances from zero but never carries a number across zero, so the left-to-right order of the two sides is unchanged.

$a < b \;\text{ and }\; k > 0 \;\Rightarrow\; ka < kb$

The direction is preserved, and the solution set is too.

Multiplying by a negative reflects both numbers through zero, and a reflection swaps their order.

$a < b \;\Rightarrow\; -b < -a$

So the symbol must reverse. The solution set is still preserved; only the direction turns around.

Multiply each side by $-2$: the left $3$ becomes $-6$ and the right $8$ becomes $-16$. Because the multiplier is negative, reverse the symbol.

$3 < 8 \;\xrightarrow{\;\times(-2)\;}\; -6 > -16$

The statement $-6 > -16$ is true, since $-16$ lies farther to the left. Keeping the symbol as $<$ gives the false claim $-6 < -16$.

Divide both sides by the positive number $a$. Dividing by a positive keeps the direction.

$ax > 5 \;\Rightarrow\; x > \frac{5}{a}$

The solution is the ray $\left(\frac{5}{a}, \infty\right)$.

Divide both sides by the negative number $a$. Dividing by a negative reverses the direction.

$ax > 5 \;\Rightarrow\; x < \frac{5}{a}$

The solution is the ray $\left(-\infty, \frac{5}{a}\right)$.

With $a = 0$ the left side $ax$ is $0$ for every $x$, so the inequality has no $x$ left in it.

$0 \cdot x > 5 \;\Rightarrow\; 0 > 5$

The statement $0 > 5$ is false, which is why this case has no solution.

The reduced statement has no variable, so its truth decides the whole problem.

$0 > -4 \text{ is true}$

Because it holds for every $x$, every real number is a solution and the solution set is all of $\mathbb{R}$.

A value solves '$P$ or $Q$' when it satisfies at least one of the pieces.

$P \text{ or } Q \;\Longleftrightarrow\; (\text{solutions of } P) \cup (\text{solutions of } Q)$

So you sweep together everything either piece allows, which is the union.

Read the chain from the middle outward: $x$ is greater than $-2$ and less than $5$.

$-2 < x < 5 \;\Longleftrightarrow\; x > -2 \;\text{ and }\; x < 5$

Its solution set is the intersection of the two rays, the band between $-2$ and $5$.

The term $2x$ is defined for every real number, so adding it can be undone by subtracting it.

$\text{add } 2x \;\xrightarrow{\;\text{undo}\;}\; \text{subtract } 2x$

A reversible step neither adds nor loses a solution, so the solution set is preserved.

The two pieces are the negative numbers and the numbers above $1$, joined by 'or', so take their union.

$x < 0 \;\text{ or }\; x > 1 \;\Longleftrightarrow\; (-\infty, 0) \cup (1, \infty)$

Each open interval matches a strict inequality, and the union symbol joins the two 'or' pieces.