Solve $|x+7| \geq 8$
A) $x \leq-15$ or $x \geq 1$
B) $x \leq-15$ and $x \geq 1$
C) $-15 \leq x \leq 1$
D) $x \geq 1$ or $x \geq-15$
Answer (1)
Split the absolute value inequality into two cases: x + 7 ≥ 8 and x + 7 ≤ − 8 .
Solve the first inequality: x + 7 ≥ 8 which gives x ≥ 1 .
Solve the second inequality: x + 7 ≤ − 8 which gives x ≤ − 15 .
Combine the solutions: x ≤ − 15 or x ≥ 1 , so the final answer is x ≤ − 15 or x ≥ 1 .
Explanation
Understanding the Problem We are given the absolute value inequality ∣ x + 7∣ ≥ 8 . Our goal is to find all values of x that satisfy this inequality. Absolute value inequalities can be solved by splitting them into two separate inequalities.
Splitting into Cases The absolute value inequality ∣ x + 7∣ ≥ 8 means that the distance between x + 7 and 0 is greater than or equal to 8. This leads to two cases:
Case 1: x + 7 ≥ 8 Case 2: x + 7 ≤ − 8
Solving Case 1 Let's solve Case 1: x + 7 ≥ 8 . To isolate x , we subtract 7 from both sides of the inequality:
x + 7 − 7 ≥ 8 − 7 x ≥ 1
Solving Case 2 Now let's solve Case 2: x + 7 ≤ − 8 . Again, we isolate x by subtracting 7 from both sides of the inequality:
x + 7 − 7 ≤ − 8 − 7 x ≤ − 15
Combining the Solutions Combining the solutions from both cases, we have x ≥ 1 or x ≤ − 15 . This means that x can be any number greater than or equal to 1, or any number less than or equal to -15.
Final Answer The solution to the inequality ∣ x + 7∣ ≥ 8 is x ≤ − 15 or x ≥ 1 . This corresponds to option A.
Examples
Absolute value inequalities are useful in various real-world scenarios. For example, consider a machine that produces parts with a specified length of 10 cm. Due to manufacturing tolerances, the actual length can vary by up to 0.5 cm. This situation can be modeled by the absolute value inequality ∣ x − 10∣ ≤ 0.5 , where x is the actual length of the part. Solving this inequality helps determine the acceptable range of lengths for the manufactured parts, ensuring they meet the required specifications. Similarly, in finance, absolute value inequalities can be used to model acceptable deviations from a target investment return.
