2006-10-22 05:48:53 · 11 answers · asked by woodrr9 1 in Science & Mathematics ➔ Mathematics
2(x-3)>4-(x+3)
2x-6>4-x-3
2x-6>1-x
+x ___+x [add x on both sides]
-------------
3X-6>1
+6__ +6 [add 6 on both sides]
-----------
3X>7
_____
3___3 [devide by 3 on both sides]
x>7/3<----Answer
any solutions of X has to be greater than 7/3
2006-10-22 07:22:25 · answer #1 · answered by Anonymous · 0⤊ 0⤋
x>7/3
2006-10-22 05:53:58 · answer #2 · answered by traveling_soldier6784 2 · 0⤊ 0⤋
x>7/3
2006-10-22 05:52:59 · answer #3 · answered by rwbblb46 4 · 0⤊ 0⤋
2(x - 3) > 4 -(x+3)
open brackets up
2x - 6 > 4 - x - 3
2x - 6 > 1 - x
move x terms to left and numbers to right (remeber to change +- when passing over the > sign)
2x + x > 1 + 6
3x > 7
x > 7/3 or 2 and 1/3
2006-10-22 05:54:17 · answer #4 · answered by rchlbsxy2 5 · 0⤊ 0⤋
2(x - 3) > 4 - (x + 3)
2x - 6 > 4 - x - 3
2x - 6 > -x + 1
3x > 7
x > (7/3)
2006-10-22 10:41:49 · answer #5 · answered by Sherman81 6 · 0⤊ 0⤋
2(x-3)>4 -(x+3)
2x-6 >4 -x-3
2x-6 > 1-x
3x>7
x>3/7
2006-10-22 05:59:25 · answer #6 · answered by ironduke8159 7 · 0⤊ 0⤋
2(x-3)-4+(x+3)>0
2x-6-4+x+3>0
3x-7>0
3x>7
x>7/3
the solution is all the numbers which are bigger than 7/3
2006-10-22 06:37:14 · answer #7 · answered by reloaded 1 · 0⤊ 0⤋
x would have to be -4, or smaller. Such as -4,-5,-6,-7, etc.
2006-10-22 05:56:21 · answer #8 · answered by Jeffrey M 2 · 0⤊ 0⤋
2x-6>4-x-3
2x-6>1-x
2x>7-x
3x>7
x>-7/3
2006-10-22 05:57:38 · answer #9 · answered by Anonymous · 0⤊ 0⤋
zero
2006-10-22 06:22:41 · answer #10 · answered by Anonymous · 0⤊ 0⤋