How do you graph and solve $|4x – 3|+ 2 <11$ ?

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Answer 1 · 2018-07-16T18:51:14

The solution is $x in (-3/2,3)$

The inequality is

$|4x-3|+2<11$

$|4x-3|-9<0$

The point to consider is

$4x-3=0$

$=>$ , $x=3/4$

There are $2$ intervals to consider

$(-oo, 3/4)$ and $(3/4,+oo)$

Therefore,

In the first interval

$-4x+3-9<0$

$=>$ , $-4x-6<0$

$=>0$ , $4x> -6$

$=>$ , $x> -3/2$

This solution belongs to the interval

In the second interval

$4x-3-9<0$

$=>$ , $4x-12<0$

$=>$ , $x<3$

This solution belongs to the interval

The solution is $x in (-3/2,3)$

graph{|4x-3|-9 [-20.27, 20.27, -10.14, 10.14]}

How do you graph and solve |4x – 3|+ 2 <11|4x – 3|+ 2 <11?