Posted by Bethany on Wednesday, July 2, 2014 at 11:48pm.
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+|5-3x|%2C+y%3D5%2C+y%3D9+for+-2+%3C+x+%3C+2
Yes I got -4/3 with 5-3x<9 and 14/3 with 5-3x>-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 absolute-value 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
-2x-8 < 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