Don't we deserve to see your approach?
Brady said:
..? Didn't I just explain that I tried this problem for a terriby long time, and I would like a clear explanation?
Brady said:
A pretty simple problem, but I'm confused nonetheless.
|x-8|<|2x+1|
Help would be greatly appreciated.
|x-8| - |2x+1| < 0
This solution is based on the fact that if a continuous function changes its sign, it might happen only in this function zeroes.
1.
Mark zeroes for each |expression| on the number line: x=8 and x=-1/2.
This two points break the whole number line into three intervals:
I. {-inf, -1/2}: -inf < x < 8
II. [-1/2, 8}: -1/2 <= x < 8
III. [8, +inf}: 8<= x < +inf.
Each expression "keeps" its sign unchanged on each interval.
Draw the number line and put the expressions' signs on the intervals
x-8: ....-...|...-...|...+...
2x+1: ...-...|...+...|...+...
Solve inequality on each region separately following the definition
|a| = -a, if a<0
|a| = a, if a>=0
I.
x-8<0 and 2x+1<0 therefore
-(x-8) - (-(2x+1)) < 0
-x + 8 + 2x + 1 < 0
x < -9 - agrees with -inf < x < 8 condition.
II.
x-8<0 and 2x+1>=0 therefore
-(x-8) - (2x+1) < 0
-x + 8 - 2x - 1 < 0
3x > 7
x > 7/3 - considering -1/2 <= x < 8 restriction
the answer on this interval
7/3 < x < 8
III.
x-8>=0 and 2x+1>=0 therefore
x - 8 - (2x+1) < 0
x - 8 - 2x - 1 < 0
x > - 9 - considering x>=8 restriction
the answer on this interval
x >= 8.
Combininig all the answers,
x < -9 or x > 7/3.
Faster way - just like Tide recommended.
Draw two graphs:
f(x) = |x-8| and g(x) = |2x+1|
(do you need help with plotting them?).
|x-8|<|2x+1| is equvalent to
graph f(x) is below graph g(x).
These two graphs intersect at x=-9 and x=7/3.
Graph f(x) is below graph g(x) left of -9 and right of 7/3.
The same answer.
