How do I solve the inequality for x and express the solution in "interval notation" for [1/(x-1)]<=(2/3)

Answer 1

1/(x-1) ≤ 2/3

subtract 2/3 for both sides to set the equations less than or equal to 0

1/(x-1) - 2/3 ≤ 0

LCD
3/(3x - 3) - (2x - 2) / (3x - 3) ≤ 0

simplify
(3 - (2x - 2)) / (3x - 3) ≤ 0

distribute
(3 - 2x + 2) / (3x - 3) ≤ 0

combine like terms
(5 - 2x) / (3x - 3) ≤ 0

set the numerator and the denominators equals to 0
5 - 2x = 0
-2x = -5
x = 5/2

3x - 3 = 0
3x = 3
x = 1

so you get x = 1 and x = 5/2

to determine which value of x that makes the equation less than or equal to 0, try to plug in some numbers

try 0
(5 - 2x) / (3x - 3) < 0
(5 - 2*0) / (3*0 - 3) 5/-3 -5/3 < 0

the statement is true, so x < 1

no try 3
(5 - 2(3)) / (3*3 - 3) ≤? 0
-1 ≤? 6
-1 ≤ 6

the statement is true so x ≥ 5/2

so the answer is:
(- infinity, 1) U [5/2 , infinity)

the 1 is not included be because the would make the equation undefined because you can't divide by a 0.

Answer 2

Multiply by x-1 to get it out of the denominator, however, this gives us two cases.

When x-1 is positive, the multiplication is okay as-is. But when x-1 is negative, the inequality has to be flipped (multiplying by a negative number flips the inequality: if a>b then -2*a < -2*b).

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case 1: x - 1 is positive:

1/(x-1) <= 2/3
1 <= 2/3(x-1)
3/2 <= x - 1
5/2 <= x
x >= 5/2

Now we also know x-1 is positive, so an additional limitation is:

x - 1 > 0
x > 1

All numbers greater than or equal to 5/2 are greater than one, so this part of the solution is just x >= 5/2.

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case 2: x - 1 is negative:

1/(x-1) <= 2/3
1 >= 2/3(x-1)
3/2 >= x - 1
5/2 >= x
x <= 5/2

Now we also know x-1 is negative, so an additional limitation is:

x - 1 < 0
x < 1

All numbers less than one are also less than or equal to 5/2, so this part of the solution is x < 1.

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Answer:

x >= 5/2
and
x < 1

There are two ranges (i.e., excluding everything from 1 inclusive to 5/2 exclusive).

Now let's test some numbers:

x = -10 is in the "x < 1" part of the solution:
1/(-10 - 1) <= 2/3
-1/11 <= 2/3 Check.

x = 3 is in the "x >= 5/2" part of the solution:
1/(3 - 1) <= 2/3
1/2 <= 2/3 Check.

x = 2 is in the excluded area between 1 and 5/2:
1/(2 - 1) <= 2/3
1 <= 2/3 Check. The inequality is false, because that number isn't part of the solution.