Jamarcus Spears

2023-03-11

How to graph and solve $|4x\u20133|+2<11$?

Anastasia Lee

Beginner2023-03-12Added 5 answers

The inequality is

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

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

The point to consider is

$4x-3=0$

$x=\frac{3}{4}$

There are $2$ intervals to consider

$(-\infty ,\frac{3}{4})$ and $(\frac{3}{4},+\infty )$

Hence,

In the first interval

$-4x+3-9<0$

$-4x-6<0$

$4x>-6$

$x>-\frac{3}{2}$

This is located in the interval.

In the second interval

$4x-3-9<0$

$4x-12<0$

$x<3$

This is located in the interval.

The answer is $x\in (-\frac{3}{2},3)$

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

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

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

The point to consider is

$4x-3=0$

$x=\frac{3}{4}$

There are $2$ intervals to consider

$(-\infty ,\frac{3}{4})$ and $(\frac{3}{4},+\infty )$

Hence,

In the first interval

$-4x+3-9<0$

$-4x-6<0$

$4x>-6$

$x>-\frac{3}{2}$

This is located in the interval.

In the second interval

$4x-3-9<0$

$4x-12<0$

$x<3$

This is located in the interval.

The answer is $x\in (-\frac{3}{2},3)$

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