\(|3x-2|≤ \frac{1}{2}\)
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 ≤ \(\frac{5}{6}\)
Attachment:
temp-2.JPG [ 24.42 KiB | Viewed 5917 times ]
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 AHope it helps!
Watch the following video to learn how to Solve Inequality + Absolute value Problems