How do you solve $| p + 8 | + 2 \leq 29$ $|p + 8| + 2\leq 29$ ?

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Answer 1 · 2017-09-24T08:19:56

see below

$| p + 8 | + 2 \leq 29$ $abs(p+8)+2<=29$

Subtract $2$ $2$ from LHS and RHS,

$| p + 8 | + 2 - 2 \leq 29 - 2$ $abs(p+8)+2-2<=29-2$

$| p + 8 | \leq 27$ $abs(p+8)<=27$

$- 27 \leq p + 8 \leq 27$ $-27<=p+8<=27$ [this pattern for "less than or equal"]

$- 27 - 8 \leq p + 8 - 8 \leq 27 - 8$ $-27-8<=p+8-8<=27-8$

$- 35 \leq p \leq 19$ $-35<=p<=19$

The interval notation is $[- 35, 19]$ $[-35,19]$

How do you solve |p+8|+2≤29|p + 8| + 2\leq 29?