Absolute Value Equations and Inequalities

9000026409

Level: 
B
Consider the following equation. \[ |2x - 4| = 5x - 7 \] Solving the equation on the intervals where it is possible to evaluate the absolute value we get equations on partial subintervals as follows. \[\begin{aligned} \text{for }x &\in (-\infty ;2)\colon &\text{for }x &\in [ 2;\infty )\colon & & & & \\ - 2x + 4 & = 5x - 7 &2x - 4 & = 5x - 7 & & & & \\ - 7x & = -11 & - 3x & = -3 & & & & \\x & = \frac{11} {7} &x & = 1 & & & & \end{aligned}\] Find the solution set of the original equation.
\(\left \{\frac{11} {7} \right \}\)
\(\left \{\frac{11} {7} ;1\right \}\)
\(\left \{1\right \}\)
\(\emptyset \)

9000027308

Level: 
A
Identify the solution set of the following inequality. \[ |2x - 1| > 5 \]
\(\left (-\infty ;-2\right )\cup \left (3;\infty \right )\)
\(\left (-\infty ;-4.5\right )\cup \left (5.5;\infty \right )\)
\(\left (1.5;\infty \right )\)
\(\left (-\infty ;0\right )\cup \left [ 5;\infty \right )\)

9000027304

Level: 
A
Identify the solution set of the following inequality. \[ |x - 1| > 10 \]
\(\left (-\infty ;-9\right )\cup \left (11;\infty \right )\)
\(\left (-\infty ;9\right )\cup \left (11;\infty \right )\)
\(\left [ 11;\infty \right )\)
\(\left (-\infty ;10\right )\cup \left [ 11;\infty \right )\)