SOLUTION: A couple of questions from my textbook that I need help with. I'm having trouble with rational inequalities. Solve the following inequalities and give your answer in interval n

Algebra ->  Inequalities -> SOLUTION: A couple of questions from my textbook that I need help with. I'm having trouble with rational inequalities. Solve the following inequalities and give your answer in interval n      Log On


   



Question 949423: A couple of questions from my textbook that I need help with. I'm having trouble with rational inequalities.
Solve the following inequalities and give your answer in interval notation.
a. -3≤ 3-5x/3 <3/8
b. 3x-5/x >0
Hint: Either both the numerator and the denominator are positive, or both are negative; consider both cases.
Thanks.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
a. -3≤ 3-5x/3 <3/8
b. 3x-5/x >0

When typing algebra, you must put parentheses around 
numerators and denominators if they contain more than 
1 letter or 1 number, like this

a. -3≤ (3-5x)/3 <3/8
b. (3x-5)/x >0

----------------------------

-3%3C=+%283-5x%29%2F3+%3C3%2F8 

Multiply thru all 3 sides by 24

-72%3C=+8%283-5x%29+%3C9

-72%3C=24-40x+%3C9

Add -24 to all 3 sides:

-72%3C-40x+%3C9

-96%3C=-40x%3C-15

Divide through by -40 which reverses
the inequalities

%28-96%29%2F%28-40%29%3E=x%3E%28-15%29%2F%28-40%29

12%2F5%3E=x%3E-3%2F8

-------------------------------

b. (3x-5)/x >0

Critical values are when numerator equas zero and when
denominator equals zero.

Numerator=0
3x-5 = 0
  3x = 5
   x = 5%2F3 = 1%262%2F3   <-- critical value

Denominator = 0
   x = 0

Put critical values on number line:

------------o------o---------------
            0     5%2F3

Test any value less than 0, say -1, by substituting
in the original:

%283x-5%29%2Fx+%3E+0

%283%5E%22%22%28-1%29-5%29%2F%28-1%29+%3E+0

%28-3-5%29%2F%28-1%29+%3E+0

%28-8%29%2F%28-1%29%3E0

8%3E0

That's true so we shade line left of 0

<===========o------o---------------
            0      5%2F3

---------------

Next test any value between 0 and 5%2F3, say 1, by substituting
in the original:

%283x-5%29%2Fx+%3E+0

%283%5E%22%22%281%29-5%29%2F%281%29+%3E+0

%283-5%29+%3E+0

-2%3E0


That's false so we do not shade line between 0 and 5%2F3. We
still have

<===========o------o---------------
            0      5%2F3

-----------------

Test any value right of 5%2F3, say 2, by substituting
in the original:

%283x-5%29%2Fx+%3E+0

%283%5E%22%22%282%29-5%29%2F%282%29+%3E+0

%286-5%29%2F2+%3E+0

%281%29%2F2%3E0

1%2F2%3E0

That's true so we shade line right of 5%2F3

S

<===========o------o==========>
            0      5%2F3

Interval notation 



Edwin