\(x+1=0 \\ \\ x=-1 \\ \\ \pmb 1^{o} \\ \\ x\in (-\infty,-1) \\ \\ -5|x+1|=10 \\ \\ 5(x+1)=10 \\ \\ 5x+5=10 \\ \\ 5x=5 \\ \\ x=1
ot\in D \\ \\ \pmb 2^{o} \\ \\ x\in [-1, +\infty) \\ \\ -5|x+1|=10 \\ \\ -5(x+1)=10 \\ \\ -5x-5=10 \\ \\ -5x=15 \\ \\ x=-3
ot\in D \\ \\ x
ot\in \Re\)
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