Inequalities & Absolute value
posted by Bethany on .
2x+4<5
53x<9
What values of x are both of the inequalities above true?
i. 2
ii. 1
iii. 1
I got 1 only and its right because I just plugged it in to find out, but is there another way of solving this perhaps algebraically? For example something like 2x+4<5 and 5? And 53x<9 or 9? The you solve for those or something? I tried that but it didn't work...

I get 4/3 < x < 1/2, so only (ii) works.
In this case, it's easy enough just to substitute in each of the values to see which works in both cases.
Always keep in mind the V shape of absolute value graphs. See
http://www.wolframalpha.com/input/?i=plot+y+%3D+2x%2B4%2C+y%3D+53x%2C+y%3D5%2C+y%3D9+for+2+%3C+x+%3C+2 
Yes I got 4/3 with 53x<9 and 14/3 with 53x>9. And then same goes for 1/2 with ...<5 and 9/2 with ....>5. I just don't get why you used 4/3<x<1/2?

For x =  2
 2 x + 4  < 5
 2 * (  2 ) + 4  < 5
  4 + 4  < 5
 0  < 5
0 < 5
Correct
 5  3 x  < 9
 5  3 * (  2 )  < 9
 5 + 6  < 9
 11  < 9
11 < 9
Not correct
1 correct solution
For x =  1
 2 x + 4  < 5
 2 * (  1 ) + 4  < 5
  2 + 4  < 5
 2  < 5
2 < 5
Correct
 5  3 x  < 9
 5  3 * (  1 )  < 9
 5 + 3 < 9
 8  < 9
8 < 9
Correct
2 correct solutions
For x = 1
 2 x + 4  < 5
 2 * 1 + 4  < 5
 2 + 4  < 5
 6  < 5
6 < 5
Not correct
 5  3 x  < 9
 5  3 * 1  < 9
 5  3  < 9
2 < 9 Correct
 5 + 6  < 9
 11  < 9
11 < 9
Not correct
1 correct solution
Answer ii 
OK. To solve absolutevalue problems, you really have to do them twice. Recall the definition of z:
z = z if z < 0
z = z if z >= 0
So, you have
2x+4<5
That means that
If 2x+4 < 0, (or, x < 2)you have
(2x+4) < 5
2x8 < 5
2x < 13
x > 13/2
But, this is not a solution, since we started out by assuming x < 2.
Or, if 2x+4 >= 0 (or, x >= 2),
2x+4 < 5
2x < 1
x < 1/2
This is ok, since 1/2 >= 2.
Now you can do the other one in the same way, and wind up with the interval I mentioned at first. 
oops  a typo. should be x > 13/2
so, the solution to the first one is 13/2 < x <= 1/2 
[10x + 8}v_2