How do you solve and graph $| 10 - 3 x | \geq 17$ $abs(10-3x)>=17$ ?

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Answer 1 · 2018-06-22T12:43:14

The solution is $x \in (- \infty, - \frac{7}{3}] \cup [9, + \infty)$ $x in (-oo,-7/3]uu [9,+oo)$

This is an inequality with absolute values

$| 10 - 3 x | \geq 17$ $|10-3x|>=17$

Therefore,

${(10-3x>=17),(-10+3x>=17):}$

$<=>$ , ${(3x<=10-17),(3x>=17+10):}$

$<=>$ , ${(x<=-7/3),(x>=9):}$

The solution is

$x in (-oo,-7/3]uu [9,+oo)$

The graph is

graph{|10-3x|-17 [-25.5, 25.83, -17.22, 8.43]}

How do you solve and graph abs(10-3x)>=17|10−3x|≥17abs(10-3x)>=17?