# Inequalities & Absolute value

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|2x+4|<5
|5-3x|<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 5-3x<9 or -9? The you solve for those or something? I tried that but it didn't work...

• Inequalities & Absolute value -

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

• Inequalities & Absolute value -

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?

• Inequalities & Absolute value -

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

• Inequalities & Absolute value -

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.

• Inequalities & Absolute value -

oops - a typo. should be x > -13/2
so, the solution to the first one is -13/2 < x <= 1/2

• Inequalities & Absolute value -

[10x + 8}v_2

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