Solve: $5|x+7|+2>8$
The answer has the form:
$(A, B)$
$(-\infty, A) \cup(B, \infty)$
Answer (1)
Isolate the absolute value term: \frac{6}{5}"> ∣ x + 7∣ > 5 6 .
Split the absolute value inequality into two cases: \frac{6}{5}"> x + 7 > 5 6 or x + 7 < − 5 6 .
Solve each inequality separately: -\frac{29}{5}"> x > − 5 29 or x < − 5 41 .
Express the solution in interval notation: ( − ∞ , − 5 41 ) ∪ ( − 5 29 , ∞ ) . The final answer is ( − ∞ , − 5 41 ) ∪ ( − 5 29 , ∞ )
Explanation
Understanding the Problem We are given the inequality 8"> 5∣ x + 7∣ + 2 > 8 . Our goal is to solve for x and express the solution in the form ( − ∞ , A ) ∪ ( B , ∞ ) . This means we're looking for all values of x that satisfy the inequality, and we expect the solution to be two separate intervals on the number line.
Isolating the Absolute Value First, we isolate the absolute value term. Subtract 2 from both sides of the inequality: 8 - 2"> 5∣ x + 7∣ + 2 − 2 > 8 − 2
6"> 5∣ x + 7∣ > 6
Dividing by 5 Next, divide both sides of the inequality by 5: \frac{6}{5}"> 5 5∣ x + 7∣ > 5 6
\frac{6}{5}"> ∣ x + 7∣ > 5 6
Splitting into Cases The inequality \frac{6}{5}"> ∣ x + 7∣ > 5 6 means that the distance between x + 7 and 0 is greater than 5 6 . This is equivalent to two separate inequalities:
\frac{6}{5}"> x + 7 > 5 6 or x + 7 < − 5 6
Solving the First Inequality Now, we solve each inequality separately. For the first inequality, subtract 7 from both sides: \frac{6}{5}"> x + 7 > 5 6
\frac{6}{5} - 7"> x > 5 6 − 7
\frac{6}{5} - \frac{35}{5}"> x > 5 6 − 5 35
\frac{6-35}{5}"> x > 5 6 − 35
-\frac{29}{5}"> x > − 5 29
-5.8"> x > − 5.8
Solving the Second Inequality For the second inequality, subtract 7 from both sides: x + 7 < − 5 6
x < − 5 6 − 7
x < − 5 6 − 5 35
x < − 5 6 + 35
x < − 5 41
x < − 8.2
Expressing the Solution The solution is the union of the two intervals: x < − 5 41 or -\frac{29}{5}"> x > − 5 29 . In interval notation, this is ( − ∞ , − 5 41 ) ∪ ( − 5 29 , ∞ ) . Therefore, A = − 5 41 = − 8.2 and B = − 5 29 = − 5.8 .
Examples
Absolute value inequalities are useful in many real-world scenarios. For example, consider a manufacturing process where a machine is set to produce parts with a specific length. Due to slight variations in the machine's operation, the actual length of the parts may vary. We can use an absolute value inequality to define an acceptable range of lengths. If the target length is L and the acceptable tolerance is T , then the actual length x must satisfy ∣ x − L ∣ < T . This ensures that all parts produced are within the specified tolerance range, maintaining quality control.
