2006-07-08 18:53:25 · 9 answers · asked by memoriesatTC 1 in Education & Reference ➔ Homework Help
no solutions for x, because -5 never > -2
2006-07-09 06:00:42 · answer #1 · answered by R@inY 3 · 3⤊ 0⤋
There are no values of x that satisfy this equation. Consider:
3x - 5 > 2(x - 1) + x
3x - 5 > 2x - 2 + x
3x - 5 > 3x - 2
-5 >-2 - this is never true.
2006-07-08 18:57:53 · answer #2 · answered by Pascal 7 · 0⤊ 0⤋
First simplify the inequality:
3x - 5 > 2x - 2 + x
3x - 5 > 3x -2
-5 > -2, which isn't true.
So, no values of x are possible.
2006-07-09 09:24:26 · answer #3 · answered by Anonymous · 0⤊ 0⤋
3x - 5 > 2(x-1) + x
First you get rid of the brackets by multiplying 2 into x-1:
3x - 5 > 2x - 2 +x
Then bring all the x groups to one side, and you'll realize you end up with 0x. Lol... you probably copied the wrong question.
2006-07-08 18:59:18 · answer #4 · answered by kaede 1 · 0⤊ 0⤋
3x-5>2x-2+x [distributive property]
3x-5>3x-2 [combining like terms]
0-5>0-2 [subtract 3x from both sides]
-5>-2 [simple math]
-5 is never greater than -2; therefore, the answer is empty set or null set or undefined.
2006-07-08 19:02:25 · answer #5 · answered by slobberknocker_usa 7 · 0⤊ 0⤋
3x-5>2(x-1)+x *x-1 by 2
3x-5>2x-2+x 2x+x
3x-5>3x-2 - 3x to both
-5>-2
this is not true
2006-07-08 19:07:23 · answer #6 · answered by isopropyl 2 · 0⤊ 0⤋
3x - 5> 2(x-1) + x
or, 3x-5>2x-2+x
or, 3x-5>3x-2
or, 3x-3x>-2+5
or, 0>3
which is not possible
2006-07-08 19:29:49 · answer #7 · answered by vishal jha 2 · 0⤊ 0⤋
Can't help you here - but thx for the points.
2006-07-08 18:56:32 · answer #8 · answered by seancanputt 2 · 0⤊ 0⤋
Empty set or { }
2006-07-08 19:03:52 · answer #9 · answered by pupa 1 · 0⤊ 0⤋