Solving Linear Inequalities: Additional Practice (Set 2)

Subtract $4x$ from both sides, and the variable cancels completely.

$4x - 1 \le 4x + 6 \;\Rightarrow\; -1 \le 6$

The leftover $-1 \le 6$ is true and free of $x$, so every real number is a solution.

Subtract $3x$ from both sides, then the constants.

$4x + 2 \le -10 \;\Rightarrow\; 4x \le -12 \;\Rightarrow\; x \le -3$

Dividing by the positive number $4$ keeps the direction.

Subtract $3$ from all three parts to get $-2 < \tfrac{x}{2} \le 1$, then multiply all three by the positive number $2$.

$-4 < x \le 2$

The left end is strict (parenthesis) and the right end is inclusive (bracket), giving $(-4, 2]$.

Subtract $4$ from both sides to get $6 \ge -2x$, then divide by $-2$ and reverse the symbol.

$\frac{6}{-2} \le \frac{-2x}{-2} \;\Rightarrow\; -3 \le x$

Reading $-3 \le x$ from the other end gives $x \ge -3$; the flip came from dividing by a negative.

Solve each piece separately: $x - 2 < -3$ gives $x < -1$, and $x - 2 > 3$ gives $x > 5$.

$x < -1 \;\text{ or }\; x > 5 \;\Rightarrow\; (-\infty, -1) \cup (5, \infty)$

The two rays do not overlap, so the solution is their union.

Subtract $x$ from both sides so the variable stays right with a positive coefficient.

$-6 > 3x + 3 \;\Rightarrow\; -9 > 3x \;\Rightarrow\; -3 > x$

Reading $-3 > x$ from the other end gives $x < -3$, with no flip needed.

Subtract $5$ from both sides to get $-9 \le -3x$, then divide by $-3$ and reverse the symbol.

$\frac{-9}{-3} \ge \frac{-3x}{-3} \;\Rightarrow\; 3 \ge x$

Reading $3 \ge x$ from the other end gives $x \le 3$; the flip came from dividing by a negative.

Subtract $2x$ from both sides so the variable moves right with a positive coefficient.

$5 > 3x - 7 \;\Rightarrow\; 12 > 3x \;\Rightarrow\; 4 > x$

Reading $4 > x$ from the other end gives $x < 4$, with no flip needed.

Solve each piece: $2x + 1 \le -3$ gives $x \le -2$, and $2x + 1 \ge 7$ gives $x \ge 3$.

$x \le -2 \;\text{ or }\; x \ge 3 \;\Rightarrow\; (-\infty, -2] \cup [3, \infty)$

Both boundaries are inclusive, so both take brackets, and the solution is the union of the two rays.

Distribute the left side to get $8x - 12 \le 8x + 5$, then subtract $8x$ from both sides.

$8x - 12 \le 8x + 5 \;\Rightarrow\; -12 \le 5 \;\text{ (true)}$

The variable cancels and the leftover $-12 \le 5$ is true, so every real number is a solution.

Combine the two conditions into one compound inequality with the smaller bound on the left.

$-2 < n \le 4$

The left end is strict (parenthesis) and the right end is inclusive (bracket), giving $(-2, 4]$.

Distribute and combine on the left: $3x + 2 - 2x = x + 2$, so $x + 2 < x + 5$. Subtract $x$ from both sides.

$x + 2 < x + 5 \;\Rightarrow\; 2 < 5 \;\text{ (true)}$

The variable cancels and the leftover $2 < 5$ is true, so every real number is a solution.