2006-09-15 19:15:31 · 10 answers · asked by Titanic lover 2 in Science & Mathematics ➔ Mathematics
Assume part 1= x, and part 2= y
Then:
x+y = 184 (i)
x/3=y/7+8 (ii)
multiply (ii) by 21
7x=3y+168
x=(3y+168)/7 (iii)
subtitute (iii) in (i)
(3y+168)/7+y=184
3y+168+7y=1288
10y=1288-168
y = 112
subtitute y value in (i)
x=184-112= 72
2006-09-15 19:28:43 · answer #1 · answered by Anonymous · 1⤊ 0⤋
Divide 184 into 2 areas such that one-0.33 of one area might exceed one-seventh of the different area by using 8. enable the two areas be ' x' and (184 - x). Then, a million/3x - a million/7(184 - x) = 8. a million/3x + a million/7x = 8 + a million/7(184) 10/21x = 8 + 26 2/7 10/21x = 34 2/7 10x = 720 x = seventy two 184 = seventy two + 112
2016-12-12 09:17:28 · answer #2 · answered by ? 4 · 0⤊ 0⤋
Let x=one part of the division of 184
Let 184-x=the other part
x/3 - (184-x)/7 = 8
7x - 3(184-x) = 168
7x - 552 + 3x = 168
10x = 168 + 552
10x = 720
x = 72
184-x=112
The two numbers are 72 and 112.
2006-09-15 19:39:18 · answer #3 · answered by fictitiousness ;-) 2 · 0⤊ 0⤋
184 = a + b
a = 184 - b
a/3 = 8 + b/7
.. mult by 21 ...
7a = 168 + 3b
7(184 - b) = 168 + 3b
1288 - 7b = 168 + 3b
1288 - 168 = 7b + 3b = 10b
1120 = 10b
112 = b
a = 184 - b = 184 - 112 = 72
a=72
b=112
check:
a/3 = 8 + b/7
72/3 = 8 + 112/7
24 = 8 + 16
24 = 24
looks right
2006-09-15 19:29:34 · answer #4 · answered by atheistforthebirthofjesus 6 · 0⤊ 0⤋
if we suppose that (x) is the first part an (y) is the second part now the question means that
(x)+(y)=184 and (1/3)(x)=(1/7)(y)+8
now we have two equation and two variable and we can solve it:
from the second equation we can deduct that:
x=3*((1/7)(y)+8)=(3/7)(y)+24
now we can put this (x) in the first equation instead of (x):
(3/7)(y)+24+(y)=184
(10/7)(y)=160
y=(7/10)160=112
so x=184-112=72
therefore the first part (x)=112 and the second part (y)=72
2006-09-15 19:34:22 · answer #5 · answered by F M 1 · 0⤊ 0⤋
184/2 = 92
one third of 92 is 92/3 = 30.67
then 184/8 = 24
one seventh of 24 is 24/7 = 3.42
2006-09-15 20:50:45 · answer #6 · answered by jaha_jaha555 1 · 0⤊ 0⤋
Let first part be x, and second part be y.
According to the question, x+y=184. let this be equation (1).
also, 1/3x-1/7y=8
=> 7x-3y=168 (taking least common multiple)
now use substitution. from equation (1), we get, x=184-y
substituting this value of x in the above obtained equation,
we get,
7(184-y)-3y=168
1288-7y-3y=168
1288-168=10y
1120=10y
or, y=1120/10=112.
substitute this value of y in equation (1)
to get x=72.
2006-09-16 01:55:31 · answer #7 · answered by noesis 2 · 0⤊ 0⤋
142 and 42
2006-09-15 19:42:27 · answer #8 · answered by ACHTUNG DE F@T@LE 1 · 0⤊ 0⤋
1/3x-1/7y=8
7x-3y=168
3x+3y=552
10x=720
x=72
y=112
2006-09-15 19:24:33 · answer #9 · answered by raj 7 · 0⤊ 0⤋
Do you have big boobies?
2006-09-15 19:22:47 · answer #10 · answered by Anonymous · 0⤊ 0⤋