SOLUTION: OK I am stuck on how to do this. Can you please explain it.The problem goes like this: Solve and graph the solution set. a) 5y+3,_< -2 or y-7_>0 b) x-9< 3x-4< x+8 PLEASE HELP

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Question 87078: OK I am stuck on how to do this. Can you please explain it.The problem goes like this: Solve and graph the solution set.
a) 5y+3,_< -2 or y-7_>0
b) x-9< 3x-4< x+8
PLEASE HELP

Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
OK I am stuck on how to do this. Can you please explain it.
The problem goes like this: Solve and graph the solution set.
a)5y + 3 < -2 or y - 7 > 0
b) x - 9 < 3x - 4 < x + 8
PLEASE HELP!!!!!!!!!

a)5y + 3 < -2 or y - 7 > 0

Solve each part:

5y + 3 < -2 or y - 7 >  0
    -3   -3       +7   +7 
-------------------------
5y     < -5 or y     >  7

    5y < -5 or y > 7

     y < -1 or y > 7

Draw three number lines with a closed circle 
at the boundary points (closed because it's 
"<" and not "<"):

y < -1                  
-----------l-------------l----------
          -1              7      

y > 7                     
-----------l-------------l----------
          -1              7  
                                                                            
y < -1 or y > 7  
-----------l-------------l----------
          -1              7

Shade the first one to the left of -1, because
"less than" is "left":

y < -1
<===========l------------l----------
           -1             7      

Shade the second one to the right of 7, because
"greater than" is "right":

y > 7
-----------l-------------l==========>
          -1             7  

Now since the word between the inequalities is "OR" 
and not "AND", we shade all parts of the final
number line if it is shaded on the first one OR if 
it is shaded on the second one.

So the final number line is:

y < -1 or y > 7  
<============l-------------l==========>
            -1             7    

Now we imagine that there is a "negative infinity"
on the far left and an "infinity" on the far right.

So the interval notation is

(-oo, -1] U [7, oo)


b) x - 9 < 3x - 4 < x + 8

Add -x + 9 to all three sides:

 x - 9 < 3x - 4 <  x + 8
-x + 9   -x + 9 < -x + 9
------------------------
     0 < 2x + 5 <     17

Add -5 to all three sides:

     0 < 2x + 5 < 17
    -5       -5   -5
------------------------
    -5 < 2x     < 12

Divide all three sides by 2,
to get only x in the middle:

    -2.5 < x < 6

That means 

-2.5 < x  AND  x < 6

which is the same as

 x > -2.5 AND x < 6


Draw three number lines with an open circle at 
the boundary points (closed because it's ">", "<" 
and not "<",">:

 x > -2.5                  
-----------o-------------o----------
         -2.5            6      

 x < 6                     
-----------o-------------o----------
         -2.5            6  
                                                                            
 x > -2.5 and x > 6  
-----------o-------------o----------
         -2.5            6

Shade the first one to the right of -2.5, because
"greater than" is "right":

 x > -2.5
-----------o========================>
         -2.5            6      

Shade the second one to the left of 6, because
"less than" is "left":

 x < 6
<========================o----------
         -2.5            6  

Now since the word between the inequalities this
time is "AND" and not "OR", we shade only the parts 
of the final number line if it is shaded on the 
first one AND shaded on the second one.

So the final number line shaded is:

 x > -2.5 and x > 6  
-----------o=============o----------
         -2.5            6

In interval notation that is 

          (-2.5, 6)

Edwin

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