solve the inequality (2x-1)/(3x-2)<=1
*<= is less than or equal to...
2007-08-26 07:56:25 · 6 answers · asked by Aerol 1 in Science & Mathematics ➔ Mathematics
2x-1<=3x-2
1<=x
x=>1
Interval: [1, ∞)
2007-08-26 08:02:20 · answer #1 · answered by de4th 4 · 0⤊ 2⤋
(2x - 1) / (3x - 2) ≤ 1
subtract 1 for both sides to set the equation less than or equao to 0
(2x - 1) / (3x - 2) - 1 ≤ 0
(2x-1) / (3x-2) - (3x-2) / (3x-2) ≤ 0
(2x - 1 - 3x + 2) / (3x - 2) ≤ 0
(-x + 1) / (3x - 2) ≤ 0
set the numerator and denominator equal to 0
-x + 1 = 0
-x = -1
x = 1
3x - 2 = 0
3x = 2
x = 2/3
so the roots are 2/3 and 1
plug in some number in to test.
try 0
(-x + 1) / (3x - 2) ≤ 0
(0+1)/(0-2) ≤ 0
1/-2 ≤ 0
the statement is true. so x < 2/3
try 2
(-2 + 1) / (6-2) ≤ 0
-1/4 ≤ 0
the statement is true. SO x ≥ 1
answer
x < 2/3 or x ≥ 1
2007-08-26 08:09:36 · answer #2 · answered by 7 · 1⤊ 0⤋
No disrespect to the people who posted them, but the first two answers fell into a trap on the very first step.
To get from (2x-1)/(3x-2)<=1 to (2x-1) <= (3x-2) you have to multiply both sides by (3x-2).
But the <= inequality only holds of (3x-2) is positive, ie if x>2/3.
If (3x-2) is negative, the inequality must be reversed to give (2x-1) >= (3x-2).
The method used in the third and fourth answers (rearranging, without multplying/dividing, to get one side to zero) is much more direct and less prone to such error.
2007-08-26 08:15:02 · answer #3 · answered by SV 5 · 0⤊ 0⤋
(2x-1)/(3x-2) -1 <=0
(2x-1-3x+2)(3x-2)<=0
(-x+1)/(3x-2)<=0 so x>=1 and x<2/3
2007-08-26 08:13:45 · answer #4 · answered by santmann2002 7 · 1⤊ 0⤋
2x^3 + 2x^2 = 4x +4 => x^3 + x^2 - 2x - 2 = 0 => x^2(x + a million) - 2(x + a million) = 0 => (x^2 - 2)(x + a million) = 0 => (x + sqrt(2))(x - sqrt(2))(x + a million) = 0 for that reason, x = sqrt(2), -sqrt(2), or -a million
2016-12-12 12:23:08 · answer #5 · answered by selders 4 · 0⤊ 0⤋
the full solution is:
(2x-1)<=(3x-2)X1
2x-1<=3x-2
2x<=3x-1
-1x<=-1
x>=1
2007-08-26 08:06:05 · answer #6 · answered by Anonymous · 0⤊ 2⤋