If |2x−7|≤8, what is the greatest possible value of |x−7|? A −1/2 B 0 C 1/2 D 15/2

Absolute Value Definition: To solve the inequality |2x−7|≤8, we need to consider the definition of absolute value, which states that |a| ≤ b implies -b ≤ a ≤ b for any real number a and non-negative real number b. So, we can write the inequality as two separate inequalities: -8 ≤ 2x−7 ≤ 8Solving Left Inequality: Now, we solve the left inequality for x: -8 ≤ 2x−7 Add 7 to both sides: -1 ≤ 2x Divide both sides by 2: -1/2 ≤ xSolving Right Inequality: Next, we solve the right inequality for x: 2x−7 ≤ 8 Add 7 to both sides: 2x ≤ 15 Divide both sides by 2: x ≤ 15/2Combining Inequalities: Combining both inequalities, we get the solution for x: -1/2 ≤ x ≤ 15/2Finding Greatest Value: 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 both endpoints: For x = -1/2: |x−7| = |-1/2−7| = |-(1/2 + 14/2)| = |-(15/2)| = 15/2 For x = 15/2: |x−7| = |15/2−7| = |15/2−14/2| = |1/2| = 1/2Comparing the two values, we see that 15/2 is greater than 1/2, so the greatest possible value of |x−7| is 15/2.

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If $|2x-7| \leq 8$ , what is the greatest possible value of $|x-7|$ ? $\newline$ (A) $-\frac{1}{2}$ $\newline$ (B) $0$ $\newline$ (C) $\frac{1}{2}$ $\newline$ (D) $\frac{15}{2}$

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Q. If

|2x-7| \leq 8

, what is the greatest possible value of

|x-7|

\newline

(A)

-\frac{1}{2}

\newline

(B)

0

\newline

(C)

\frac{1}{2}

\newline

(D)

\frac{15}{2}

Absolute Value Definition: To solve the inequality $|2x-7|\leq 8$ , we need to consider the definition of absolute value, which states that $|a| \leq b$ implies $-b \leq a \leq b$ for any real number $a$ and non-negative real number $b$ . $\newline$ So, we can write the inequality as two separate inequalities: $\newline$ $-8 \leq 2x-7 \leq 8$
Solving Left Inequality: Now, we solve the left inequality for $x$ : $-8 \leq 2x-7$ Add $7$ to both sides: $-1 \leq 2x$ Divide both sides by $2$ : $-\frac{1}{2} \leq x$
Solving Right Inequality: Next, we solve the right inequality for $x$ : $2x−7 \leq 8$ Add $7$ to both sides: $2x \leq 15$ Divide both sides by $2$ : $x \leq \frac{15}{2}$
Combining Inequalities: Combining both inequalities, we get the solution for $x$ : $-\frac{1}{2} \leq x \leq \frac{15}{2}$
Finding Greatest Value: 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 $x$ . $\newline$ We calculate $|x-7|$ at both endpoints: $\newline$ For $x = -\frac{1}{2}$ : $|x-7| = \left| -\frac{1}{2}-7 \right| = \left| -\left(\frac{1}{2} + \frac{14}{2}\right) \right| = \left| -\frac{15}{2} \right| = \frac{15}{2}$ $\newline$ For $x = \frac{15}{2}$ : $|x-7| = \left| \frac{15}{2}-7 \right| = \left| \frac{15}{2}-\frac{14}{2} \right| = \left| \frac{1}{2} \right| = \frac{1}{2}$
Finding Greatest Value: 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 both endpoints: For $x = -\frac{1}{2}$ : $|x-7| = \left| -\frac{1}{2}-7 \right| = \left| -\left(\frac{1}{2} + \frac{14}{2}\right) \right| = \left| -\frac{15}{2} \right| = \frac{15}{2}$ For $x = \frac{15}{2}$ : $|x-7| = \left|\frac{15}{2}-7\right| = \left|\frac{15}{2}-\frac{14}{2}\right| = \left|\frac{1}{2}\right| = \frac{1}{2}$ Comparing the two values, we see that $\frac{15}{2}$ is greater than $x$ $0$ , so the greatest possible value of $|x-7|$ is $\frac{15}{2}$ .