Absolute value equations and inequalities

9000026403

Level: 
B
To solve the equation with absolute value using interval method we have to divide the domain of the equation by the zero points of the subexpressions in the absolute value. Find all these points. \[ |x + 1| + |2x - 1| = 3 \]
\(-1,\ \frac{1} {2}\)
\(- 3\)
\(1,\ -\frac{1} {2}\)
\(0\)

9000026404

Level: 
B
To solve the equation with absolute value using interval method we have to divide the domain of the equation by the zero points of the subexpressions in the absolute value. Find all these points. \[ 2|x - 2| + |2 - x| = 1 + |x| \]
\(2,\ 0\)
\(-2,\ 2,\ 0\)
\(-1,\ 2\)
\(-1,\ 2,\ 0\)

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.
\(\left \{\frac{11} {7} \right \}\)
\(\left \{\frac{11} {7} ;1\right \}\)
\(\left \{1\right \}\)
\(\emptyset \)