I need help please.......?

Question

not sure how to do this, I have never seen it with a number after the sign only a zero.....


x+1/x-1 <_ 3

this problem reads as x+1 over x-1 is less than or equal to 3

the answer given is x<1 or x greater than or equal to 2

let me clarify that usually this problem is done by setting x+1 and x-1 to equal zero and then using a number line and test interval pts..........however, there is usually a zero rather than a three there.

this type of problem doesn't combine the left side...........
the correct answer given for this was:

x<1 or x (>or= sign)2

not x/x =1 this is incorrect this is a different type of problem

this is inequalities problem, where you set the equations on the left to equal zero and then use a number line broken into intervals to solve.

Answer 1

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

=> combine fractions on the left hand side

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

=> multiply by x-1 on both sides

CASE x-1 > 0

Then, we don't have to switch the inequality:

x(x-1) + 1 <= 3(x-1)

=>

x^2 - x + 1 <= 3x - 3

=>

x^2 -4x + 4 <=0

=>

(x-2)^2 <= 0, which can never happen, unless x = 2. Note that x-1 = 1 > 0, as we assumed, so this solution is OK.

CASE x-1 < 0

Now we have to switch the inequality:

x(x-1) + 1 >= 3(x-1)

=>

x^2 - x + 1 >= 3x - 3

=>

(x-2)^2 >= 0, which is always true. But remember, we assumed that x-1 < 0 here. So we can only have x < 1.

Thus, the complete solution set is:

{x | x < 1 or x = 2}

EDIT

Try plugging in x = 5 into the inequality and see how you won't get a true statement. x >= 2 is incorrect! x = 2 works, but nothing greater than 2.

Answer 2

x/(x-1) + 1/(x-1) < = 3 clear the fractions

x^2 -x + (x-1) <= 3(x-1)

x^2 - x + x - 1 <= 3x - 3

x^2 -3x + 2 <= 0 factor

(x-1)(x-2) <= 0 critical #'s that yield zero are 1 and 2

but 1 must be excluded because it makes the fraction undef.

So try a test number in the three regions defined by the intervals less than 1 and between 1 and 2 and greater than or equal to 2

You discover x<1 or x>= 2 will give the desired results

Answer 3

This problem is asking you to provide a solution set to prove that the equation is valid. For example if X = 2 then 2+1+3 and 2-1 =1 then : 3/1=3 and 3 is less than or equal to 3. However if you use 1for example....1+1=2 and 1-4= -3 and 2/-3= -0.66.. so any negative number is auomatically less than 1. You get the idea.....

Answer 4

y = x+1/(x-1)= < 3
There is a discontinuity at x =1 so there is a vertical asymptote at that point. There is also a horizontal asymptote at y = 1 -So y <3 for x=>2 and x<1
(x+1)/(x-1) =< 3
x+1 =< 3(x-1)
x+1=< 3x-3
4 =< 2x
2= x >= 2