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Solve for $p$ . $\newline$ $|p + 1| < 9$ $\newline$ Write a compound inequality like $1 < x < 3$ or like $x < 1$ or $x > 3$ . Use integers, proper fractions, or improper fractions in simplest form. $\newline$ ______

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Solve absolute value inequalities

Full solution

Q. Solve for

p

.

\newline

|p + 1| < 9

\newline

Write a compound inequality like

1 < x < 3

or like

x < 1

or

x > 3

. Use integers, proper fractions, or improper fractions in simplest form.

\newline

______

Given Inequality: We are given the inequality $|p + 1| < 9$ . We need to consider two cases for the absolute value: one where the expression inside the absolute value is positive, and one where it is negative.
Case $1$ : Non-negative Expression: First, let's consider the case where the expression inside the absolute value is non-negative, which means $p + 1$ is already positive. We can remove the absolute value signs: $\newline$ $p + 1 < 9$ $\newline$ Now, we solve for $p$ by subtracting $1$ from both sides: $\newline$ $p + 1 - 1 < 9 - 1$ $\newline$ $p < 8$
Case $2$ : Negative Expression: Next, we consider the case where the expression inside the absolute value is negative, which means we need to take the opposite of the expression inside to remove the absolute value signs: $\newline$ $- (p + 1) < 9$ $\newline$ Now, we distribute the negative sign: $\newline$ $-p - 1 < 9$ $\newline$ We solve for $p$ by adding $1$ to both sides and then multiplying by $-1$ to get $p$ alone: $\newline$ $-p - 1 + 1 < 9 + 1$ $\newline$ $-p < 10$ $\newline$ Multiplying by $-1$ (and remembering to reverse the inequality sign): $\newline$ $p > -10$
Combining Cases: Combining the two cases, we get a compound inequality: $\newline$ $-10 < p < 8$ $\newline$ This means $p$ is greater than $-10$ and less than $8$ .

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Question

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\newline

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\newline

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\newline

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