In Exercises solveabsvalequfirst - solveabsvalequlast, solve the equation.
\(|x| = 6\)

\(x = -6\) or \(x=6\)
\(|3t-1| = 10\)

\(t = -3\) or \(t= \frac {11}{3}\)
\(|4-w| = 7\)

\(w = -3\) or \(w= 11\)
\(4 - |y| = 3\)

\(y = -1\) or \(y= 1\)
\(2|5m+1| - 3 = 0\)

\(m=-\frac {1}{2}\) or \(m= \frac {1}{10}\)
\(|7x-1| + 2 = 0\)

No solution
\(\frac {5 - |x|}{2} = 1\)

\(x=-3\) or \(x= 3\)
\(\frac {2}{3} |5-2w| - \frac {1}{2} = 5\)

\(w = -\frac {13}{8}\) or \(w= \frac {53}{8}\)
\(|3t - \sqrt {2}| + 4 = 6\)

\(t = \frac {\sqrt {2} \pm 2}{3}\)
\(\frac {|2v+1| - 3}{4} = \frac {1}{2} - |2v+1|\)

\(v = -1\) or \(v = 0\)
\(|2x+1| = \frac {|2x+1| - 3}{2}\)

No solution
\(\frac {|3-2y|+ 4}{2} = 2 - |3-2y|\)

\(y = \answer {\frac {3}{2}}\)

\(|3t - 2| = |2t + 7|\)

\(t = -1\) or \(t = 9\)
\(|3x+1| = |4x|\)

\(x = -\frac {1}{7}\) or \(x = 1\)
\(|1-\sqrt {2} y| = |y+1|\)

\(y = 0\) or \(y = \frac {2}{\sqrt {2} - 1}\)
\(|4-x| - |x+2| = 0\)

\(x = \answer {1}\)

\(|2-5z| = 5 |z+1|\)

\(z = \answer {-\frac {3}{10}}\)

\(\sqrt {3}|w-1| = 2|w+1|\)

\(w = \frac {\sqrt {3} \pm 2}{\sqrt {3} \mp 2}\)
In Exercises solveinequabsquadfirst - solveinequabsquadlast, solve the inequality. Write your answer using interval notation.
\(|3x - 5| \leq 4\)

\(\left [\frac {1}{3}, 3\right ]\)
\(|7t + 2| > 10\)

\(\left (-\infty , -\frac {12}{7} \right ) \cup \left (\frac {8}{7}, \infty \right )\)
\(|2w+1| - 5 < 0\)

\((-3,2)\)
\(|2-y| - 4 \geq -3\)

\((-\infty ,1] \cup [3,\infty )\)
\(|3z+5| + 2 < 1\)

No solution
\(2|7-v| +4 > 1\)

\((-\infty , \infty )\)
\(3 - |x+\sqrt {5}| < -3\)

\((-\infty , -6-\sqrt {5}) \cup (6-\sqrt {5}, \infty )\)
\(|5t| \leq |t|+3\)

\(\left [ -\frac {3}{4}, \frac {3}{4}\right ]\)
\(|w-3| < |3-w|\)

No solution
\(2 \leq |4-y| < 7\)

\((-3,2] \cup [6,11)\)
\(1 < |2w - 9| \leq 3\)

\([3, 4) \cup (5, 6]\)
\(3 > 2|\sqrt {3} - x| > 1\)

\(\left (\frac {2 \sqrt {3} - 3}{2}, \frac {2 \sqrt {3} - 1}{2} \right ) \cup \left (\frac {2 \sqrt {3} +1}{2}, \frac {2 \sqrt {3} +3}{2} \right )\)
With help from your classmates, solve:

  1. \(|5 - |2x-3|| = 4\)

    \(x = -3\), or \(x = 1\), or \(x = 2\), or \(x = 6\)

  2. \(|5 - |2x-3|| < 4\)

    \((-3,1) \cup (2,6)\)