2006-12-03 10:11:17 · 12 answers · asked by hallaatcha2010 1 in Science & Mathematics ➔ Mathematics
- 15c + 7c + 8c = 3 - 1
0c = 2
c = 0
So there are no possible solutions.
2006-12-04 16:16:55 · answer #1 · answered by arpita 5 · 0⤊ 0⤋
C=2
2006-12-03 10:19:22 · answer #2 · answered by chrismango13 3 · 0⤊ 0⤋
bring all the terms involving c on one side and all numbers on the other side.
=>-15c+7c+8c=3-1
=>0=2
=>no value of c can satisfy the equation i.e no solution!
2006-12-03 18:35:08 · answer #3 · answered by Anonymous · 0⤊ 0⤋
first of all you take the -15 and add 7 to it giving you
-8c +1=3 -8c Then you have to get all of your variables on one side so subtract 1 giving you -8c=2 -8c then again move your variable add 8c to both sides and the positive 8 brings the -8 to 0
so it is 0=2 which is not possible. So there are no possible solutions.
2006-12-03 10:26:17 · answer #4 · answered by harley_rose85 1 · 0⤊ 0⤋
if we solve throw logic c=2 but as the equation is not solvent so no solution that is -15c+7c+1is not equal to 3-8c
2006-12-04 01:57:05 · answer #5 · answered by mani k 2 · 0⤊ 0⤋
-15c+7c+1=3-8c
-8c+1=3-8c
0c+1=3
c=2
i think
2006-12-03 12:45:41 · answer #6 · answered by ♥KiYa♥ 3 · 0⤊ 0⤋
-15c+7c+1=3-8c
-15c+7c+8c=3-1
-15c+15c=2
0=2
2006-12-06 17:37:01 · answer #7 · answered by sam 3 · 0⤊ 0⤋
- 15c + 7c + 8c = 3 - 1
0c = 2
c = 0
There are no solutions for c...sorry!!
2006-12-03 18:55:52 · answer #8 · answered by mad_integer 3 · 0⤊ 0⤋
c(-15+7)+1=3-8c
-8c+1=3-8c
-8c+8c=3-1
0.c=2
c=2/0
therefore it cannot be defined
2006-12-03 14:07:30 · answer #9 · answered by s madhavan 1 · 0⤊ 0⤋
that's undefined cause they cancel out to make 1=3? :/
2006-12-03 10:22:06 · answer #10 · answered by Roxanne 3 · 0⤊ 0⤋