10a+275b=343
275a+9625b=10455
2006-07-23 03:35:51 · 7 answers · asked by Murtada A 1 in Science & Mathematics ➔ Mathematics
hey listen "ibhinit" is wrong here is the answer:
10a +275b = 343
10a = 343 -275b
a= (343 -275b)/10 = 34.3 -27.5b
so by substitution of a in the second equation, but first, let's divide all the 2nd one by 5, so that it would be simpler =
55a +1925b =2091
so: 55(34.3 -27.5b) + 1925b = 2091
1886.5 -1512.5b +1925b = 2091
412.5b = 204.5
b= 0.4958 = 409/825
so: 10a + 275(409/825) = 343
10a + 136 1/3 = 343
10a = 206 2/3
a = 20.666666 = 20 2/3
2006-07-23 03:47:24 · answer #1 · answered by Anonymous · 0⤊ 0⤋
10a + 275b = 343
275a + 9625b = 10455
10a + 275b = 343
55a + 1925b = 2091
Multiply the top by -55 and the bottom by 10
-550a - 15125b = -18865
550a + 19250b = 20910
4125b = 2045
b = 409/825
10a + 275b = 343
10a + 275(409/825) = 343
10a + (4499/33) = 343
10a + (409/3) = 343
30a + 409 = 1029
30a = 620
a = (62/3)
a = (62/3) or 20.67
b = (409/825) or .495757575
2006-07-23 05:25:54 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
Well if you mean to find out what a and b equal it's:
a= 20+2/3 b=409/825
or you can write it like this
a=20.66666667 b=0.495757575
:P hope i helped
Oww and this is the right answer :P
2006-07-23 03:55:24 · answer #3 · answered by PookAir 1 · 0⤊ 0⤋
10a + 275b = 343
275a+9625b=10455
350a + 9625b = 12005
275a + 9625b = 10455
------------------------------
75a = 1550
a = 20 2/3
b = 409/825
2006-07-23 04:12:48 · answer #4 · answered by menezes_dean 2 · 0⤊ 0⤋
from first equation a= 343-275b
then in second equation replacing value of a {275(343-275b)} obtain the value of b. after this u can very easily find the value of a.
b = 83870/66000 = 1.2707
a = 0.61425
2006-07-23 03:44:40 · answer #5 · answered by Anonymous · 0⤊ 0⤋
a = 20â666666....
b = 0â49575757.....
10a + 275b =
10(20â66666...) + 275(0â49575757...)= 343
275a + 9625b =
275(20â66666.....) + 9625(0â49575757.....) = 10455
2006-07-23 08:27:35 · answer #6 · answered by Brenmore 5 · 0⤊ 0⤋
how old are you?
2006-07-23 03:39:56 · answer #7 · answered by nastea 2 · 0⤊ 0⤋