If x>3, then x^2+2x-3/x^2-9

Question

1.If x>3, then x^2+2x-3/x^2-9

 

2. solve the inequality. then graph the solution set |2x-2|+4>12

Answer 1

For question 1, I assume the directions wanted you to simplify this   x² +2x-3

                                                                                                  x² -9

 

First you would need to factor both the top (numerator) and bottom(denominator) of the equation then see what cancels out. Notice the denominator is a difference of squares

 

(x+3)(x-1)            the (x+3) will cancel out leaving an answer of   x-1

(x+3)(x-3)                                                                                x-3

 

 

For question two, remember you cannot remove an absolute value sign if there is anything else on that side of the equation. You will need to move the 4 to the other side and then remove the absolute value sign. 

 

The general rule for absoulte value is that if you start off with an absolute value greater than a number, you use this rule     |u|>c    becomes  u>c   and u< -c

 

so for question two      

 

|2x-2| +4 > 12

          -4      -4    first move the 4 to the other side

 

|2x-2| > 8 now use the rule to get      2x-2 >8     and   2x-2<-8     solve both equations

                                                

2x-2>8             2x-2<-8

  +2  +2              +2   +2

 

2x> 10              2x<-6

 

2x>10               2x<-6     Divide both sides of each equation by 2 to get x by itself

2    2                  2    2

 

x>5                   x<-3

 

To graph this, you would have a number line with an open circle at -3 and the arrow would point to the left, you would have another open circle at 5 and the arrow would point to the right

 

FYI  if you get an absolute value that is in the form   |u|<c   it would become   -c<u<c    

 

also if you multiply or divide as your last step in getting x alone, don't forget to switch the direction of the sign