How do you solve this?

Question

1/|3x+2| is greater than or equal to 1 The Second Equation is |(2x+5)/3|<1 If the answers could be more of the principles and the process than just the answer that would be great! Thanks in advance

Answer 1

The first inequality is given as

1 <= 1/ |3x +2| For the inequality to be true the denominator must be none negative and it must be less than 1.

This means the inequlaity is true iff
0 < 3x + 2 <= 1 now simplify

-2 < 3x <= -1

-2/3 < x <= -1/3. This is now easily verifiable. Check three values one value to the left of -2/3, one value between -2/3 and -1/3 and one value to the right of -1/3. You should also check the -2/3 and -1/3 to see how these values react in the inequality. By doing this, you will always get them correct.

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the second inequalit is given as

|(2x + 5)/3| < 1 Since the number 3 is always positive it can be removed from the abs brackets and mult to both sides of the inequality

|2x + 5| < 3 Now to remove the abs brckts the inequality goes to

- 3 < 2x + 5 < 3 simplifying the inequal

- 8 < 2x < -2

- 4 < x < -1 Now, as above, test points on the left, right and in between this inequality. You will see that these are the only values that make the original inequality true.

The whole trick to inequalities is THINK first!

Good luck

Answer 2

There are two equations
1/(3x+2)>=1 and 1/(3x+2)<=-1
1/(3x+2) -1 =( -3x-1)/( 3x+2) >0 -2/3 2)3¡3x+3)/(3x+2)<=0 -1<=x<-2/3
so the solution is -1<=x<= -1/3 except x= -2/3

Answer 3

1/|3x+2| >= 1
1 >= |3x+2|

Solve for both +(3x+2) and -(3x+2)

1) 1 >= 3x+2
-1 >= 3x
-1/3 >= x

2) 1 >= -3x-2
3 >= -3x
-1 <= x

So -1 <= x <= -1/3, except x = -2/3 (that would make the denominator = 0)


|(2x+5)/3| < 1

Solve for both +[(2x+5)/3] and -[(2x+5)/3]

(2x+5)/3 < 1
2x+5 < 3
2x < -2
x < -1

(2x+5)/(-3) < 1
2x+5 > -3
2x > -8
x > -4

So -4 < x < -1