Question

question 3 of 10 (1 point) | question attempt: 1 of unlimited (a) |y+1|=0 (b) |y+1|>0 (c) |y+1|<0 part 1 of 4 (a) the solution set for |y+1|=0 is {-1}. part: 1 / 4 part 2 of 4 (b) the solution set in interval notation for |y+1|>0 is

Explanation:

Step1: Recall absolute value property

The absolute value of a number is always non - negative. That is, for any real number \(x\), \(|x|\geq0\), and \(|x| = 0\) if and only if \(x = 0\). For the inequality \(|y + 1|>0\), this means that \(y+1
eq0\) (because when \(y + 1=0\), \(|y + 1| = 0\) which does not satisfy the strict inequality \(|y+1|>0\)).

Step2: Solve for \(y\)

If \(y + 1
eq0\), then \(y
eq - 1\). In interval notation, the set of all real numbers except \(y=-1\) can be written as the union of two intervals: \((-\infty,-1)\cup(-1,\infty)\).

Answer:

\((-\infty, - 1)\cup(-1,\infty)\)