1. ## Inequalities help!

Solve the ff inequalities:
need solution set
4.) 2 > -3 - 3x >= -7
5.) 4/x - 3 - 2/x + 7 > 0
6.) 1/(x+1) < 2/(3x - 1)

2 > -3 - 3x >= -7
If what I'm about to do bothers you, please feel free to do each step on the two inequalities:
2 > -3 - 3x and -3 - 3x >= 7

So
2 > -3 - 3x >= -7

Add 3 to all three "sides":
2 + 3 > 3 - 3 - 3x >= 3 - 7

5 > -3x >= -4

Divide all three "sides" by -3. Since we are dividing by a negative quantity, don't forget to switch the > to < and the >= to <=!
5/(-3) < -3x/(-3) <= -4/(-3)

-5/3 < x <= 4/3

-Dan

4/x - 3 - 2/x + 7 > 0
4/x - 3 - 2/x + 7 > 0

Simplify the LHS:
2/x + 4 > 0

Add the two numbers on the LHS:
2/x + 4*x/x > 0

(2 + 4x)/x > 0

Now we wish to look for "critical points." The denominator is equal to zero at x = 0, so this is a cp. Next, the numerator is zero at x = -1/2, so this is a cp.

Break the real number line into the following pieces:
(-infinity, -1/2), (-1/2, 0), and (0, infinity)

For which of these three regions is the inequality satisfied?
(-infinity, -1/2): (2 + 4x)/x > 0 (Check!)
(-1/2, 0) : (2 + 4x)/x < 0 (No.)
(0, infinity): (2 + 4x)/x > 0 (Check!)
(If we have <= and/or >= we also need to check the cp's themselves.)

So the solution set is x < -1/2 and x > 0.

-Dan

1/(x+1) < 2/(3x - 1)
1/(x+1) < 2/(3x-1)

This uses the same method as the previous problem.

1/(x+1) < 2/(3x-1)

1/(x+1) - 2/(3x-1) < 0

[(3x-1) - 2(x+1)]/[(x+1)(3x-1)] < 0

(x-3)/[(x+1)(3x-1)] < 0

This has cp's at x = 3, x = 1/3, and x = -1.
(-infinity, -1): (x-3)/[(x+1)(3x-1)] < 0 (Check!)
(-1, 1/3): (x-3)/[(x+1)(3x-1)] > 0 (No.)
(1/3, 3): (x-3)/[(x+1)(3x-1)] < 0 (Check!)
(3, infinity): (x-3)/[(x+1)(3x-1)] > 0 (No.)

Thus the solution set is: x < -1 and 1/3 < x < 3.

Now, if we did this problem the intuitive way:
1/(x+1) < 2/(3x-1)

Multiply both sides by (x+1):
1 < 2(x+1)/(3x-1)

Multiply both sides by (3x-1):
3x - 1 < 2(x + 1)
3x - 1 < 2x + 2
x < 3

Note that this did not give the correct answer. This is because we are not saying whether or not (x+1) or (3x-1) is negative when we multiply by them. Always be careful of this!

-Dan