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solve the inequality ||x-5|+2|≤3

Sagot :

ferrybukantoro

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}

semsee45

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.

  • To solve an equation containing an absolute value:
  • Isolate the absolute value on one side of the equation.
  • Set the contents of the absolute value equal to both the positive and negative value of the number on the other side of the equation.
  • Solve both equations.

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]

Learn more about inequalities here:

https://brainly.com/question/27947009

https://brainly.com/question/27743925

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