Solving Linear Inequalities: Practice

Subtract $2$ from all three parts to get $-7 < -x \le 1$. Now divide all three by $-1$, reversing both symbols, and the bounds swap ends.

$7 > x \ge -1 \;\Rightarrow\; -1 \le x < 7$

The left end is inclusive (bracket) and the right end is strict (parenthesis), giving $[-1, 7)$.

Subtract the matching variable term from both sides of each and read the leftover. Only one leaves a false statement.

$2x + 3 < 2x - 1 \;\Rightarrow\; 3 < -1 \;\text{ (false)}$

The other three each reduce to a true statement, so they hold for all real numbers; only $3 < -1$ is impossible.

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

$7 > 2x - 5 \;\Rightarrow\; 12 > 2x \;\Rightarrow\; 6 > x$

Reading $6 > x$ from the other end gives $x < 6$, with no flip needed.

Subtract $1$ to get $-\tfrac{2}{3}x \le 4$, then multiply both sides by $-\tfrac{3}{2}$ and reverse the symbol.

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

The flip comes from multiplying by a negative number.

Distribute and combine like terms: $5x - 3x + 6 = 2x + 6$, so $2x + 6 < 2$. Subtract $6$, then divide by the positive number $2$.

$2x < -4 \;\Rightarrow\; x < -2$

No flip is needed, because the final division is by a positive number.

Distribute on both sides, then collect the variable on the left.

$6x - 8 \ge 5x - 5 \;\Rightarrow\; x - 8 \ge -5 \;\Rightarrow\; x \ge 3$

No division by a negative occurs, so the symbol stays as $\ge$.

The left boundary $-2$ has an open circle, so it is excluded and takes a parenthesis. The right boundary $1$ has a filled circle, so it is included and takes a bracket.

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

The shaded band is the numbers between them.

Subtract $2$ from both sides to get $\tfrac{x}{-4} > 1$, then multiply both sides by $-4$ and reverse the symbol.

$-4 \cdot \frac{x}{-4} < -4 \cdot 1 \;\Rightarrow\; x < -4$

The flip comes from multiplying by a negative number.

Distribute both products and combine: $3x + 6 - 2x + 2 = x + 8$, so $x + 8 \le 0$.

$x + 8 \le 0 \;\Rightarrow\; x \le -8$

Subtracting $8$ does not change the direction.

Add $x$ to both sides so the variable moves right with a positive coefficient.

$9 \ge 3x - 3 \;\Rightarrow\; 12 \ge 3x \;\Rightarrow\; 4 \ge x$

Reading $4 \ge x$ from the other end gives $x \le 4$, with no flip needed.

Distribute the left side to get $6x - 6 > 6x - 4$, then subtract $6x$ from both sides.

$6x - 6 > 6x - 4 \;\Rightarrow\; -6 > -4 \;\text{ (false)}$

The variable cancels and the leftover $-6 > -4$ is false, so there is no solution.

Distribute and combine on the left: $5 - 2x - 6 = -1 - 2x$, so $-1 - 2x \ge x - 4$. Add $2x$ to both sides to keep the variable positive.

$-1 \ge 3x - 4 \;\Rightarrow\; 3 \ge 3x \;\Rightarrow\; 1 \ge x$

Reading from the other end gives $x \le 1$, with no flip needed.