Question

Solve the inequality with absolute value
$|3x+1|\le 2x+3$

• A. $[\frac{4}{5},2]$
• B. $(-\infty ,+\infty )$
• C. no solutions
• D. all real numbers

Answer 1

The correct option is A $[\frac{4}{5},2]$

Since $|x|=x$ and $-x$, therefore,

$|3x+1|<=2x+3$ may be calculated as:

$3x+1<=2x+3$ and $-(3x+1)<=2x+3$

$3x-2x<=3-1$ and $-3x-1<=2x+3$

$x<=2$ and $-3x-2x<=3+1$

$x<=2$ and $-5x<=4$

$x<=2$ and $x>=-\frac{4}{5}$

Therefore, $x$ lies between $[-\frac{4}{5}2]$.

Hence option A is correct.