Question 1235: Show me how to work this.
|3x-4| > 11
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! Case (i) If 3x - 4 >= 0,
then |3x-4| > 11 converts to
3x - 4 > 11
or 3x > 11 + 4 = 15,
So, x > 5...(*)
But,by 3x - 4 >= 0, or 3x >= 4 or x >= 4/3...(1)
Hence, the numbers x > 5 satisfying the given condition (1)
Case (ii) If 3x - 4 < 0,
then |3x-4| > 11 converts to
-(3x - 4) > 11 , or -3x + 4 > 11,
or -3x > 11 - 4 = 7,
Dividing both sides by -3(have to change the direction of the inequality)
We have x < -7/3 ...(**)
But,by 3x - 4 < 0, or 3x < 4 or x < 4/3...(2)
We see that the numbers x < -7/3 satisfying the given condition (2)
Answer: x > 5 or x < -7/3
The solution set is (-oo, -7/3) U (5,+oo)
Another way: USe |ax -b| > c (note a > 0,c >=0) is equivalent to
x > (b+c)/a or x < (c -b)/a
Now a = 3, b = 4, c = 11 and we have
x > 5 or x < -7/3
Moreover, if c < 0, |ax -b| > c is true for all x and the solution
set is R(all real numbers)
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