2006-06-28 20:39:39 · 11 answers · asked by joy 1 in Education & Reference ➔ Homework Help
Answer: 2a
group all the a's together (5a+4a-7a=-2a),
then all the b's together (2b+b-3b=0)
2006-06-28 20:43:46 · answer #1 · answered by Athenart 2 · 0⤊ 0⤋
5a+2b-7a-3b+4a+b = 2a
2006-06-28 20:47:01 · answer #2 · answered by jayant p 1 · 0⤊ 0⤋
we could see... stack em a+ 3b - c= a million -4a- 2b +5c=sixteen 7a+10b+6c=-15 p.c.. 2 and get rid of a variable (in case you won't be in a position to by potential of merely including multiply each and each term in an equation by potential of a style to make it nicely suited) as an occasion: 4(a+3b-c)=a million then upload and elimenate the variable 4a+12b-4c=4 -4a-2b+5c=sixteen and you get 10b+c=20 Repeat with yet another 2 equations... -7(a+3b-c=a million) 7a+10b+6c= -15 you get -7a-21b+7c=-7 7a+10b+6c=-15 upload em and you get -11b+13c=-22 Now get rid of a variable from the two new equations... -13(10b+c=20) -11b+13c=-22 you get -130b-13c=-260 -11b+13c=-22 upload and you get -141b=-282 Didvide the two aspects by potential of -141 to get *b=2* Plug it into considered one of the two equations with 2 variables 10b+c=20 10(2)+c=20 20+c=20 Subtract 20 from the two aspects to get *c=0* Then plug the two in to between the 1st 3 equations a+3b-c=a million a+3(2)-0=a million a+6=a million subtract a million from the two aspects to get *a=-5* so your very final answer a=-5 b=2 c=0
2016-10-31 21:51:56 · answer #3 · answered by ? 4 · 0⤊ 0⤋
5a+4a-7a+2b+1b-3b
9a-7a+3b-3b
2a+0
2a
2006-06-28 20:48:15 · answer #4 · answered by Anonymous · 0⤊ 0⤋
It's 2a
2006-06-28 20:44:54 · answer #5 · answered by alia_vahed 3 · 0⤊ 0⤋
2a
2006-07-05 20:11:08 · answer #6 · answered by bugzaper 3 · 0⤊ 0⤋
2a
2006-06-28 21:16:54 · answer #7 · answered by brattydoll 2 · 0⤊ 0⤋
2a
2006-06-28 21:08:37 · answer #8 · answered by Riddle 1 · 0⤊ 0⤋
2a
2006-06-28 21:02:40 · answer #9 · answered by xiiaogal 1 · 0⤊ 0⤋
2a
2006-06-28 20:43:47 · answer #10 · answered by bReAd-WiNnEr 3 · 0⤊ 0⤋