[(3x-6)/(4x-8)]<=1. Solve; use interval notation?

(3x - 6)/(4x - 8) ≤ 1
[3(x - 2)]/[4(x - 2)] ≤ 1
3/4 ≤ 1  and  x - 2 ≠ 0   (because 0/0 is undefined)

For all values of x, 3/4 < 1
Hence, -∞ < x < ∞  and  x ≠ 2
i.e. -∞ < x < 2 or 2 < x < ∞

The answer in interval notation: x ∊ (-∞, 2) or (2, ∞)

(3x - 6)/(4x - 8) ≤ 1

We need to multiply both sides by (4x - 8)² to make sure we have a positive value and the inequality stays the same.

so, (3x - 6)(4x - 8) ≤ (4x - 8)²

i.e. (4x - 8)² - (3x - 6)(4x - 8) ≥ 0 

so, (4x - 8)[(4x - 8) - (3x - 6)]  ≥ 0

so, (4x - 8)(x - 2) ≥ 0

i.e. (x - 2)(x - 2) ≥ 0

or, (x - 2)² ≥ 0....true for all values of x

As we cannot have x = 2, the solution would be (-∞, 2) ∪ (2, ∞)

:)>  

[(3x-6)/(4x-8)] ≤ 1
[3(x-2)]/[4(x-2)] ≤ 1
(3/4) ≤ 1
True for all values of x.

The value of the rational expression has to be considered in the limit as x→2 since the expression evaluates to the indeterminate value 0/0. But, since that limit exists and is equal to 3/4, the solution set also includes x = 2

If x = 2 then (3x-6)/(4x-8) = 0/0 is indefinite.
So x = 2 is not a solution.

If x < 2 then 4x-8 < 0, so
(3x-6)/(4x-8) ≦ 1
3x-6 ≧ 4x-8
2 ≧ x
But x < 2 must be satisfied, so the solution is x < 2.

If x > 2 then 4x-8 > 0, so
(3x-6)/(4x-8) ≦ 1
3x-6 ≦ 4x-8
2 ≦ x
But x > 2 must be satisfied, so the solution is x > 2.

Therefore, the solution is x < 2 or x > 2.
Using interval notation, (-inf,2) ∪ (2,inf).

 [(3x - 6)/(4x - 8)] ≤ 1
Solutions:
2 > x > 2

[(3x - 6)/(4x - 8)] ≤ 1

[(3x - 6)/(4x - 8)] - 1 ≤ 0

[(3x - 6) - (4x - 8)]/(4x - 8) ≤ 0

(3x - 6 - 4x + 8)/(4x - 8) ≤ 0

(- x + 2)/(4x - 8) ≤ 0

(- x + 2)/[4.(x - 2)] ≤ 0 → you know that: 4 > 0

(- x + 2)/(x - 2) ≤ 0

- (x - 2)/(x - 2) ≤ 0

- 1 ≤ 0 ← wrong → no solution

[(3x - 6)/(4x - 8)] <= 1
3x - 6 <= 4x - 8
x >= 2
2 < x > 2

3x-6
------ ≤ 1 ⇒
4x-8

// Multiply both sides by (4x-8)
// to get rid of fraction
3x-6 ≤ 4x-8

// Combine like-terms: Subtract
// 3x from both sides
3x-3x - 6 ≤ 4x-3x - 8
-6 ≤ x - 8

// Add 8 to both sides
-6 + 8 ≤ x - 8 + 8
2 ≤ x

Solution in interval notation is [2,+∞).........ANS