Jan 30, 2020 Thread starter #1 karush Well-known member Jan 31, 2012 2,653 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 $-7 \le y \le 1$ which is c hopefully Last edited: Jan 30, 2020
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 $-7 \le y \le 1$ which is c hopefully
Jan 30, 2020 #2 skeeter Well-known member MHB Math Helper Mar 1, 2012 643 $|y+3| \le 4 \implies -4 \le y+3 \le 4 \implies -7 \le y \le 1$
Jan 31, 2020 Thread starter #3 karush Well-known member Jan 31, 2012 2,653 That was quick.. Doesn't that assume y is positive
Jan 31, 2020 #4 skeeter Well-known member MHB Math Helper Mar 1, 2012 643 karush said: That was quick.. Doesn't that assume y is positive Click to expand... what does the inequality, ${\color{red}-7 \le y} \le 1$, tell you about the possible signs for $y$? also, see attached graph ... Attachments abs_inequality.jpg 32 KB Views: 10
karush said: That was quick.. Doesn't that assume y is positive Click to expand... what does the inequality, ${\color{red}-7 \le y} \le 1$, tell you about the possible signs for $y$? also, see attached graph ...
Jan 31, 2020 #5 skeeter Well-known member MHB Math Helper Mar 1, 2012 643 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, $y+3 < 0$ $-(y+3) \le 4 \implies y+3 \ge -4 \implies y \ge -7$
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, $y+3 < 0$ $-(y+3) \le 4 \implies y+3 \ge -4 \implies y \ge -7$