It + 3I is called absolut value which means the value of t + 3 is always positive
So we will solve two inequalities
[tex]\begin{gathered} t+3\leq2 \\ -(t+3)\leq2 \end{gathered}[/tex]Let us solve the first one
Subtract from both sides 3
[tex]t+3-3\leq2-3[/tex][tex]t\leq-1[/tex]Now we will solve the second one
We multiply the bracket by (-)
[tex]-t-3\leq2[/tex]Add 3 to both sides
[tex]\begin{gathered} -t-3+3\leq2+3 \\ -t\leq5 \end{gathered}[/tex]We will divide both sides by -1 the coefficient of t, but we must reverse the sign of the inequality because 2 < 3 if we divide both sides by -1 it will be -2 < -3 which is wrong -2 > -3, so
When you multiply or divide an inequality by negative number you must reverse the sign of the inequality
[tex]\begin{gathered} -1(-t)\ge-1(5) \\ t\ge-5 \end{gathered}[/tex]Now we will write the both inequalities togather
[tex]-5\leq t\leq-1[/tex]The minimum value is -5
The maximum value is -1