*gre.al.9 GRE Exam Inequality with modulus or absolute value

[karush]

​

given
$|y+3|\le 4$
we don't know if y is plus or negative so
$y+3\le 4 \Rightarrow y\le 1$
and
$-(y+3)\le 4$
reverse the inequality
$ y+3 \ge -4$
then isolate y
$y \ge -7$
the interval is
$-7 \le y \le 1$

 

[skeeter]

$|y+3| \le 4 \implies -4 \le y+3 \le 4 \implies -7 \le y \le 1$

 

[karush]

That was quick..
Doesn't that assume y is positive

 

[skeeter]

That was quick..
Doesn't that assume y is positive

what does the inequality, ${\color{red}-7 \le y} \le 1$, tell you about the possible signs for $y$?

also, see attached graph ...

 

[skeeter]

definition of absolute value ...

$|\text{whatever}| = \left\{\begin{matrix}
\text{whatever}, & \text{if whatever}\ge 0\\
-(\text{whatever}), & \text{if whatever}< 0
\end{matrix}\right.$

therefore ...

$|y+3| = \left\{\begin{matrix}
y+3 \, , &\text{if }y+3 \ge 0 \\
-(y+3) & \text{if }y+3<0
\end{matrix}\right.$

$|y+3| \le 4$

case 1, $y+3 \ge 0$

$y+3 \le 4 \implies y \le 1$

case 2, $y+3 < 0$

$-(y+3) \le 4 \implies y+3 \ge -4 \implies y \ge -7$