8,000X + 6,000Y = 680,000
2,000X + 10,000Y = 680,000
2006-11-21 02:45:50 · 7 answers · asked by Funtravel 1 in Science & Mathematics ➔ Mathematics
8000X + 6000Y = 680000 (1)
2000X + 10000Y = 680000 --> 8000X + 40000Y = 2720000 (2)
Take (1) from (2)
34000Y = 2040000
Y = 60
2000X + 10000x60 = 680000
2000X + 600000 = 680000
2000X = 80000
X = 40
X = 40, Y = 60
2006-11-21 02:54:07 · answer #1 · answered by Tom :: Athier than Thou 6 · 1⤊ 0⤋
8,000X + 6,000Y = 2,000X +10,000Y
6,000X = 4,000Y
3X = 2Y
Since both equations equal 680,000, both equations equal each other. Thus you get 8,000X + 6,000Y = 2,000X +10,000Y
Next, you bring the X to one side of the equation and the Y to the other side by subtracting 2,000X from 8,000X to get 6,000X on one side and on the other side you subtract 6,000Y from 10,000Y to get 4,000Y. Thus you get 6,000X = 4,000Y
Finally, you simplify the equation to get 3X = 2Y. (by dividing both sides by 2,000)
oops i didnt solve the question look to the answer below for values for x and y. But with the equation 3X = 2Y, you can substitute it into the first equation 8,000X + 6,000Y = 680,000 so it becomes (2/3)8,000Y + 6,000Y = 680,000. Thus Y = 60. Since 3X=2Y, 3X = 120 thus X = 40.
2006-11-21 02:50:00 · answer #2 · answered by holeymoley 1 · 2⤊ 0⤋
8,000X + 6,000Y = 680,000.....Eq 1
2,000X + 10,000Y = 680,000.....Eq 2
Consider Eq 1
8,000X + 6,000Y = 680,000
2000(4X + 3Y) = 680000
4X + 3Y = 680000/2000
4X + 3Y = 340
4X = 340 - 3Y
X = (340 - 3Y)/4
Substitute the derived value of X in Eq 2
2,000X + 10,000Y = 680,000
2000[(340 - 3Y)/4] + 10000Y = 680000
500(340 - 3Y) + 10000Y = 680000
170000 - 1500Y + 10000Y = 680000
8500Y = 680000-170000
8500Y = 510000
Y = 510000/8500
Y = 5100/85
Y = 60
Substitute the value of Y in Eq 1
8,000X + 6,000Y = 680,000
8000X + 6000*60 = 680000
8000X + 360000 = 680000
8000X = 680000-360000
8000X = 320000
X = 320000/8000
X = 40
X = 40, Y = 60 is the solution
2006-11-21 02:58:49 · answer #3 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 1⤋
easy ;-)
first divide by 1'000 everywhere, it's a waste of 0's
(1) 8x+6y=680
(2) 2x+10y=680
then divide by 2 also, it's more elegant
(1b) 4x+3y=340
(2b) x+5y=340
then subtract 4 times the second equation from the 1st, the x's will cancel out, you'll have:
(4x-4x) + (3y-20y) = 340 - 1'360
so
17y = 1'020
so y = 60
plug this into (1):
4x + 3*60 = 340
4x = 160
x = 40
so there you have it, x=40, y=60
hope this helps
2006-11-21 04:25:32 · answer #4 · answered by AntoineBachmann 5 · 0⤊ 0⤋
8,000X + 6,000Y = 2,000X +10,000Y
6,000X = 4,000Y
Y = (3/2)X
Then substitute back into one of the original problems
8000X + 6000 (3/2)X = 680,000
17000X = 680,000
X = 40 (and if Y = (3/2)X)) Y = 60
2006-11-21 02:59:00 · answer #5 · answered by Bill S 2 · 0⤊ 0⤋
divide all equations by 2000,we have
4x+3y=340....1
x+5y=340.......2
from eqn.2
x=340-5y
substitute in eqn.1
4[340-5y]+3y=340
17y=3*340
y=60
x=340-300=40
2006-11-21 02:55:41 · answer #6 · answered by openpsychy 6 · 1⤊ 0⤋
This has infinitely many solutions
2006-11-21 02:54:49 · answer #7 · answered by Paritosh Vasava 3 · 0⤊ 1⤋