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Step-by-step explanation:
||x-5|+2|≤3
-3 ≤ (|x-5| +2) ≤ 3
-5 ≤ |x-5| ≤ 1
|x-5| ≥ -5 => not required
|x-5| ≤ 1
-1 ≤ x-5 ≤ 1
4 ≤ x ≤ 6
the Solutions:
{4≤ x ≤ 6}
Answer:
[tex]4\leq x\leq 6[/tex]
Step-by-step explanation:
Given inequality:
[tex]||x-5|+2|\leq 3[/tex]
The absolute value of a number is its positive numerical value.
Case 1: (x - 5) is positive
[tex]\begin{aligned} \implies |x-5+2| & \leq 3\\ |x-3| & \leq 3\end{aligned}[/tex]
[tex]\textsf{Apply absolute rule}: \quad \textsf{If }|u| \leq a\:\textsf{ when } \: a > 0,\: \textsf{then }-a \leq u \leq a[/tex]
[tex]\implies -3\leq x-3\leq 3[/tex]
Therefore:
[tex]-3\leq x-3 \implies 0\leq x[/tex]
[tex]x-3\leq 3 \implies x\leq 6[/tex]
Merge the overlapping intervals:
[tex]\implies 0\leq x\leq 6[/tex]
Case 2: (x - 5) is negative
[tex]\begin{aligned} \implies |-(x-5)+2| & \leq 3 \\ |-x+5+2| & \leq 3\\ |-x+7| & \leq 3\end{aligned}[/tex]
[tex]\textsf{Apply absolute rule}: \quad \textsf{If }|u| \leq a\:\textsf{ when } \: a > 0,\: \textsf{then }-a \leq u \leq a[/tex]
[tex]\implies -3\leq -x+7\leq 3[/tex]
Therefore:
[tex]-3\leq -x+7 \implies 10\geq x[/tex]
[tex]-x+7\leq 3 \implies x\geq 4[/tex]
Merge the overlapping intervals:
[tex]\implies 4\leq x\leq 10[/tex]
Finally, merge the overlapping intervals for case 1 and 2:
[tex]\implies 4\leq x\leq 6[/tex]
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