Solving Linear Inequalities: Additional Practice (Set 1)

Subtract $4$ from both sides to get $-3x > 15$, then divide by $-3$ and reverse the symbol.

$\frac{-3x}{-3} < \frac{15}{-3} \;\Rightarrow\; x < -5$

The flip comes from dividing both sides by a negative number.

Distribute the leading negative to get $-x + 5 < 2x - 1$, then add $x$ to both sides to keep the variable positive.

$5 < 3x - 1 \;\Rightarrow\; 6 < 3x \;\Rightarrow\; 2 < x$

Reading $2 < x$ from the other end gives $x > 2$, with no flip needed.

Subtract $3x$ from both sides, and the variable cancels.

$3x - 5 \ge 3x + 1 \;\Rightarrow\; -5 \ge 1 \;\text{ (false)}$

The leftover $-5 \ge 1$ is false, and no value of $x$ can fix it, so there is no solution.

Distribute on both sides, then subtract $2x$ so the variable moves right with a positive coefficient.

$2x + 8 > 3x - 3 \;\Rightarrow\; 11 > x$

Reading $11 > x$ from the other end gives $x < 11$, with no flip needed.

The circle on $1$ is filled, so $1$ is included, which means an inclusive symbol. The shading runs left toward the smaller numbers, so the relation is 'less than or equal to'.

$\text{filled circle on } 1, \text{ shade left} \;\Rightarrow\; x \le 1$

An open circle would give $x < 1$, and shading right would give a 'greater' relation.

Solve each piece: $x + 3 \le 1$ gives $x \le -2$, and $x - 4 \ge 0$ gives $x \ge 4$.

$x \le -2 \;\text{ or }\; x \ge 4 \;\Rightarrow\; (-\infty, -2] \cup [4, \infty)$

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

Subtract $3$ from both sides to get $-5x \le -15$, then divide by $-5$ and reverse the symbol.

$\frac{-5x}{-5} \ge \frac{-15}{-5} \;\Rightarrow\; x \ge 3$

The flip is required because the divisor is negative; a negative divided by a negative is positive.

Subtract $1$ from all three parts to get $-4 \le -x < 3$. Now divide all three by $-1$, reversing both symbols, and the bounds swap ends.

$4 \ge x > -3 \;\Rightarrow\; -3 < x \le 4$

The left end is strict (parenthesis) and the right end is inclusive (bracket), giving $(-3, 4]$.

Substitute $x = -1$ into the left side and compare with the right.

$3 - 4(-1) = 7, \qquad 7 \le 9 \;\checkmark$

Since $7 \le 9$ is true, $x = -1$ is a solution.

Subtract $8$ from both sides to get $-\tfrac{x}{3} \ge -2$, then multiply both sides by $-3$ and reverse the symbol.

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

The flip comes from multiplying by a negative number.

Both boundaries have filled circles, so both are included and take brackets.

$\text{filled at } -2, \text{ filled at } 1 \;\Rightarrow\; [-2, 1]$

The shaded band is every number between $-2$ and $1$, including both ends.

Multiply all three parts by the positive number $2$, then add $1$, then divide all three by $3$.

$-4 < 3x - 1 \le 8 \;\Rightarrow\; -3 < 3x \le 9 \;\Rightarrow\; -1 < x \le 3$

The left end is strict (parenthesis) and the right end is inclusive (bracket), giving $(-1, 3]$.