Honors Algebra 2: Extra Practice on Compound Inequalities
- Class: Honors Algebra 2
- Author: Peter Atlas
- Text: Algebra and Trigonometry: Structure and Method, Brown
Solve the following:
- \(3x + 5 < 7x - 1 \leq 12x - 10\)
Solution
\( \displaystyle x \geq \frac{9}{5}\)
- \(2x+3 > 6x - 4\text{ or }5x + 9 < 15x - 8\)
Solution
all real numbers
- \( \displaystyle 2x + 1 \leq \frac{3 - 6x}{4} < \frac{5x}{3} + 2\)
Solution
\( \displaystyle -\frac{15}{38} < x \leq -\frac{1}{14}\)
- \(0.2x - 1 < 6\text{ or }1.5 - 3x \geq 4\)
Solution
\( x < 35 \)
- \(4x + 10 > 12\text{ and }3x - 5 \geq 16\)
Solution
\( x \geq 7\)
- \(7x + 1 < 15\text{ and }12x + 10 > 42\)
Solution
\( \emptyset \)
- \(8 \left| 10x - 6 \right| - 16 < 24\)
Solution
\( \displaystyle \frac{1}{10} < x < \frac{11}{10}\)
- \(-5\left| x + 20 \right| - 15 > -35\)
Solution
\( -24 < x < -16 \)
- \(\displaystyle \left| \frac{2x + 3}{8}\right| \geq 2\)
Solution
\( \displaystyle x \leq -\frac{19}{2}\text{ or }x \geq \frac{23}{2}\)
- \(\left|x + 5\right| + \left|x - 7\right| > 10\)
Solution
all real numbers
- \(\left|2x + 4\right| - \left|6x + 24\right| > 48\)
Solution
\( \emptyset \)