Q. If ∣2x−7∣≤8, what is the greatest possible value of ∣x−7∣?(A) −12(B) 0(C) 12(D) 152
Understand Absolute Value Definition: We are given the inequality ∣2x−7∣≤8. To solve this, we need to consider the definition of absolute value, which states that ∣a∣≤b implies −b≤a≤b. We apply this to our inequality.
Separate into Two Inequalities: We rewrite the inequality as two separate inequalities: −8≤2x−7≤8.
Solve Left Inequality: Now we solve for x in both inequalities. Starting with the left inequality: −8≤2x−7.Add 7 to both sides to isolate the term with x: −8+7≤2x.
Solve Right Inequality: This simplifies to −1≤2x. Now divide both sides by 2 to solve for x: −21≤x.
Combine Both Inequalities: Now we solve the right inequality: 2x−7≤8. Add 7 to both sides: 2x≤15.
Find Greatest ∣x−7∣ Value: Divide both sides by 2 to solve for x: x≤215.
Calculate |x\(-7| ext{ at Endpoints: Combining both inequalities, we have } -\frac{1}{2} \leq x \leq \frac{15}{2}. \text{This is the range of values } x \text{ can take.}
Identify Calculation Mistake: Now we need to find the greatest possible value of ∣x−7∣. To do this, we need to consider the distance of x from 7 on the number line. The greatest distance will occur at the endpoints of the interval for x.
Identify Calculation Mistake: Now we need to find the greatest possible value of ∣x−7∣. To do this, we need to consider the distance of x from 7 on the number line. The greatest distance will occur at the endpoints of the interval for x.We calculate ∣x−7∣ at the endpoints of the interval. First, for x=−21: ∣(−21)−7∣=∣−7.5∣=7.5.
Identify Calculation Mistake: Now we need to find the greatest possible value of ∣x−7∣. To do this, we need to consider the distance of x from 7 on the number line. The greatest distance will occur at the endpoints of the interval for x.We calculate ∣x−7∣ at the endpoints of the interval. First, for x=−21: ∣(−21)−7∣=∣−7.5∣=7.5.Now for x=215: ∣(215)−7∣=∣7.5∣=7.5.
Identify Calculation Mistake: Now we need to find the greatest possible value of ∣x−7∣. To do this, we need to consider the distance of x from 7 on the number line. The greatest distance will occur at the endpoints of the interval for x.We calculate ∣x−7∣ at the endpoints of the interval. First, for x=−21: ∣(−21)−7∣=∣−7.5∣=7.5.Now for x=215: ∣(215)−7∣=∣7.5∣=7.5.Both endpoints give us the same value for ∣x−7∣, which is x0. However, this value is not listed in the answer choices. We must have made a mistake in our calculations.
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