Question 14: Minimum value of (x ) satisfying \( {\left| {x + \frac{2}{3}} \right| + 2 = 2\frac{1}{3}}\)

\(\begin{array}{*{20}{l}} {\left| {x + \frac{2}{3}} \right| + 2 = 2\frac{1}{3}}\\ {\left| {x + \frac{2}{3}} \right| = \frac{7}{3} – 2}\\ {\left| {x + \frac{2}{3}} \ right| = \frac{1}{3}} \end{array}\)

+ TH1:

\(\begin{array}{*{20}{l}} {x + \frac{2}{3} = \frac{1}{3}}\\ {x = \frac{1}{3} – \frac{2}{3}}\\ {x = \frac{{ – 1}}{3}} \end{array}\)

+ TH2:

\(\begin{array}{*{20}{l}} {x + \frac{2}{3} = \frac{{ – 1}}{3}}\\ {x = \frac{{ – 1}}{3} – \frac{2}{3}}\\ {x = – 1} \end{array}\)

So x=−1 or \( x = \frac{{ – 1}}{3}\)

Or the smallest value of x that satisfies the problem is −1

The answer to choose is: GET