-gre.al.9 absolute value domain

[karush]

Solve for y: $\quad |y+3|\le 4$
a.$\quad y \le 1$
b.$\quad y\ge 7$
c.$\quad -7\le y\le1$
d. $\quad -1\le y\le7$
e. $\quad -7\ge y \ge 1$

Ok I think this could be solved by observation but is risky to do so...

Spoiler: possible solution

since y can be either plus or minus y would be between 2 -7\ge y \le 1 thus a and b are not answers

\begin{array}{l|l}
(y+3)=4&-(y+3)=4\\
y=4-3&-y=4+3\\
y=1&y=-7
\end{array}
thus
$-7\le y \le 1$
which is c.

 

[castor28]

[sp]
The distance between $y$ and $-3$ is at most $4$. That means $y$ is between $-3-4=-7$ and $-3+4=1$.
[/sp]

 

[HOI]

$|y+ 3|\le 4$ is equivalent to $-4\le y+ 3\le 4$. Subtracting 4 from each part:
$-7\le y \le 1$.

 

[karush]

well that's a lot quicker
mahalo btw how would you rate this problem easy, medium, hard

 

[skeeter]

easy, if you know $|x| \le a \implies -a \le x \le a$ for $a \ge 0$

 

[karush]

$|y+ 3|\le 4$ is equivalent to $-4\le y+ 3\le 4$. Subtracting 4 from each part:
$-7\le y \le 1$.

mahalo

 

[karush]

easy, if you know $|x| \le a \implies -a \le x \le a$ for $a \ge 0$

Mahalo