Solve for t. |-t| > 3 Write a compound inequality like 1 3. Use integers, proper fractions, or improper fractions in simplest form. ______

Define Inequality: We have the inequality: |-t| > 3 First, we need to solve for |-t|. |-t| > 3 means that the absolute value of -t is greater than 3.Split into Cases: Since the absolute value of a number is always non-negative, we can split the inequality into two cases based on the definition of absolute value: 1. -t > 3, when t is negative. 2. -t 3, we solve for t by multiplying both sides by -1, remembering to reverse the inequality sign: -t > 3 t 3 (after multiplying by -1)Combine Cases: Combining both cases, we get the compound inequality: t 3 This is the solution to the original inequality |-t| > 3.

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Solve for $t$ . $\newline$ |-t| > 3 $\newline$ 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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Algebra 2

Solve absolute value inequalities

Full solution

Q. Solve for

t

\newline

|-t| > 3

\newline

Write a compound inequality like

1 < x < 3

or like

x < 1

x > 3

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

\newline

______

Define Inequality: We have the inequality: $\newline$ |-t| > 3 $\newline$ First, we need to solve for $|-t|$ . $\newline$ |-t| > 3 means that the absolute value of $-t$ is greater than $3$ .
Split into Cases: Since the absolute value of a number is always non-negative, we can split the inequality into two cases based on the definition of absolute value: $\newline$ $1$ . -t > 3, when $t$ is negative. $\newline$ $2$ . -t < -3, when $t$ is positive (because the inside of the absolute value, $-t$ , must be less than $-3$ to make the absolute value greater than $3$ ).
Case $1$ : Negative t: For the first case, -t > 3, we solve for t by multiplying both sides by $-1$ , remembering to reverse the inequality sign: $\newline$ -t > 3 $\newline$ t < -3 (after multiplying by $-1$ )
Case $2$ : Positive $t$ : For the second case, -t < -3, we also solve for $t$ by multiplying both sides by $-1$ , again reversing the inequality sign: $\newline$ -t < -3 $\newline$ t > 3 (after multiplying by $-1$ )
Combine Cases: Combining both cases, we get the compound inequality: $\newline$ t < -3 or t > 3 $\newline$ This is the solution to the original inequality |-t| > 3.