We'll solve the equation, expressing first the
modulus.
Case 1)
l 3x-2 l = 3x
- 2 for 3x-2 >= 0
3x =<
2
x =< 2/3
Now, we'll
solve the equation:
5(3x-2) =
10
3x-2 = 2
3x =
4
x = 4/3
Since x = 4/3 does
belong to the interval of admissible values,[2/3, +infinite], we'll accept
it.
Case 2)
l 3x-2 l = -3x + 2
for
3x-2 < 0
3x<2
x<2/3
Now,
we'll solve the equation:
5(-3x+2) =
10
-3x+2 = 2
-3x =
0
x = 0
Since x = 0 does
belong to the interval of admissible values, (-infinite, 2/3), we'll accept
it.
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