Solve for x: |3x-2|≤ 1/2

Question

Solve for x:
 |3x-2|≤ \frac{1}{2}

A. ( \frac{1}{2}, \frac{5}{6} )
B. ( \frac{3}{2}, \frac{6}{5} )
C. ( \frac{5}{2}, \frac{3}{4} )
D. ( \frac{3}{7}, \frac{4}{9} )
E. ( \frac{6}{7}, \frac{4}{3} )

rajatchopra1994 · Accepted Answer

Explanation: |3x-2| ≤ 1/2 Opening the mode With +ve Sign: 3x-2 ≤ 1/2 x ≤ 5/6......a Opening mode with -ve Sign: -3x+2 ≤ 1/2 1/2≤x......b therefore, From a & b We get the limit value of x: ( \frac{1}{2}, \frac{5}{6} ) Answer is A

Please provide kudos, if you find my explanation good enough

karnavora1 · Answer

Solve for x:
 |3x−2|≤\frac{1}{2}

Just open it with a positive sign and you will get
 3x-2≤\frac{1}{2} 
 6x-4≤1 
 6x≤5 
 x≤\frac{5}{6}

Now if you notice the answers there is only 1 option with  \frac{5}{6}  and hence you don't need to solve further.
Answer is A.

BrushMyQuant · Answer

|3x-2|≤ \frac{1}{2} We will have to take two cases Case 1 : Whatever is inside the modulus is >= 0 => 3x-2 ≥ 0 => 3x ≥ 2 => x ≥ \frac{2}{3} ~ 0.67 => |3x-2| = 3x-2 (as |X| = X when X >= 0) => 3x-2 ≤ \frac{1}{2} => 3x ≤ 2 + \frac{1}{2} => 3x ≤ \frac{4+1}{2} => x ≤ \frac{5}{2*3} => x ≤ \frac{5}{6} But the condition was x ≥ \frac{2}{3} So, the solution will be the intersection of the two solution (As shown in the image below) => \frac{2}{3} ≤ x ≤ \frac{5}{6} temp-2.JPG Please note that both the points in Option A \frac{2}{3} and \frac{5}{6} satisfy the condition so we don't need to solve further. But I am finishing the problem to complete the solution for similar problems. Case 2 : Whatever is inside the modulus is < 0 => 3x-2 < 0 => 3x < 2 => x < \frac{2}{3} => |3x-2| = -(3x-2) (as |X| = -X when X < 0) => -(3x-2) ≤ \frac{1}{2} => -3x + 2 ≤ \frac{1}{2} => -3x ≤ \frac{1}{2} - 2 => -3x ≤ \frac{1-4}{2} => -3x ≤ \frac{-3}{2} => x ≥ \frac{-3}{2*-3} (multiplying both the sides will -3 will reverse the sign of the inequality) => x ≥ 2 But condition was x < \frac{2}{3} and x is not < \frac{2}{3} => NO SOLUTION in this case => \frac{2}{3} ≤ x ≤ \frac{5}{6} is the solution Clearly \frac{2}{3} and \frac{5}{6} satisfy the condition So, Answer will be A Hope it helps! Watch the following video to learn how to Solve Inequality + Absolute value Problems https://www.youtube.com/watch?v=236auvH8agc

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Solve for x: |3x-2|≤ 1/2

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Solve for x: |3x-2|≤ 1/2

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