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Question 628625: |3 + 2x| > |4 - x|
Solve and graph.
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! I imagine that you would know how to solve this if if was something like:
|3 + 2x| > 9
You would rewrite this as:
3 + 2x > 9 or 3 + 2x < -9 ("or" because the absolute value is "greater than")
and then solve the compound inequality.
Well that's exactly how we're going to start
|3 + 2x| > |4 - x|
The fact that the right side is an absolute value does not significantly change things. So we start with:
3 + 2x > |4 - x| or 3 + 2x < -|4 - x|
Now we repeat the process for the remaining absolute values! But first, let's get them on the left side. (To save time I'm not going to show you how. I hope you can figure out what I did.)
|4 - x| < 3 + 2x or |4 - x| < -3 + (-2x)
Now we rewrite each of these (using "and" since both inequalities say an absolute value is "less than"):
(4 - x < 3 + 2x and 4 - x > -(3 + 2x)) or (4 - x < -3 + (-2x) and 4 - x > -(-3 + -2x))
Now we solve this big mess. Each of the 4 inequalities are not that hard. I'll solve each one separately and then put them back together with the and's and or.
4 - x < 3 + 2x
Add x
4 < 3 + 3x
Subtract 3
1 < 3x
Divide by 3
1/3 < x
4 - x > -(3 + 2x))
4 - x > -3 + (-2x)
Add 2x:
4 + x > -3
Subtract 4:
x > -7
4 - x < -3 + (-2x)
Add 2x:
4 + x < -3
Subtract 4L
x < -7
4 - x > -(-3 + -2x)
4 - x > 3 + 2x
Add x:
4 > 3 + 3x
Subtract 3:
1 > 3x
Divide by 3:
1/3 > x
Replacing each of the 4 inequalities of
(4 - x < 3 + 2x and 4 - x > -(3 + 2x)) or (4 - x < -3 + (-2x) and 4 - x > -(-3 + -2x))
with their solutions we have:
(1/3 < x and x > -7) or (x < -7 and 1/3 > x)
Now let's analyze the results. As I hope you know, you should always read inequalities from where the variable is. (This means reading it right-to-left if the variable is on the right.) The two inequalities
(1/3 < x and x > -7)
say x is greater that something. The "and" between them means that both must be true. The only way x and be greater than both 1/3 and -7 is if x is greater than 1/3. So we can replace
(1/3 < x and x > -7)
with
(1/3 < x)
Analyzing (x < -7 and 1/3 > x) in the same way we find that both inequalities say x is less than something. And the only way x can be less than both 1/3 and -7 is if x is less than -7. So we can replace
(x < -7 and 1/3 > x)
with
(x < -7)
Our simplified solution is now
(1/3 < x)
Analyzing (x < -7 and 1/3 > x) in the same way we find that both inequalities say x is less than something. And the only way x can be less than both 1/3 and -7 is if x is less than -7. So we can replace
(x < -7 and 1/3 > x)
with just
(x < -7)
So our simplified solution is
(1/3 < x) or (x < -7)
Unfortunately I can't figure out how to get algebra.com's graphing software to show you a number line graph of inequalities. As you Might know, with "or" graphs you just plot the two inequalities onto the same number line. So just plot both x > 1/3 and x < -7 on the same number line and that will be the proper graph of (1/3 < x) or (x < -7).
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