Expert mathmeticians needed
4,194,304/x + 1 = y
3,555,443.2/x + 2 = y
I need to find such an x and y where x is the same in both, and y is the same in both
any ideas guys?
[170 byte] By [
pouncer] at [2007-11-20 11:31:13]

# 1 Re: Expert mathmeticians needed
its a standard linear equation of the form Ax + C = y.
you are given with 2 equation of above form.
just subtract 2 equations to get equation of form Ax + C = 0 from which you can get x value i.e -C\A (in this ex).
then substitute the value of x in any equation and get 'y'
# 2 Re: Expert mathmeticians needed
its a standard linear equation of the form Ax + C = y.
you are given with 2 equation of above form.
just subtract 2 equations to get equation of form Ax + C = 0 from which you can get x value i.e -C\A (in this ex).
then substitute the value of x in any equation and get 'y'
a1x + b1y + c1=0 (1)
a2x + b2y + c2=0 (2)
from (1)
x = (-c1 - b1y)/a1
put in (2)
a2(-c1 - b1y)/a1 +b2y + c2 = 0
-a2c1/a1 - a2b1y/a1 + b2y + c2 = 0
-a2c1 - a2b1y + a1b2 y + a1c2 = 0
y(a1b2 - a2b1) = a2c1 - a1c2
y = (a2c1 - a1c2) / (a1b2 - a2b1)
can you now compute x?
cilu at 2007-11-9 13:03:24 >

# 3 Re: Expert mathmeticians needed
I always liked resolving this with matrices:
|a1 b1| = |c1|
|a2 b2| |c2|
d = a1b2 - a2b1
then for x replace the matrix C in the first column:
|c1 b1|
|c2 b2|
and dx = c1b2 - c2b1
x = dx/d = (c1b2 - c2b1) / (a1b2 - a2b1)
to compute y, replace C in the second column:
|a1 c1|
|a2 c2|
and dy = a1c2 - a2c1
y = dy / d = (a1c2 - a2c1) / (a1b2 - a2b1)
It's easier this way to compute solve systems of N variables.
cilu at 2007-11-9 13:04:22 >

# 4 Re: Expert mathmeticians needed
It is not a linear equation... meaning it is not an equation to a straight line... solve the equation using substitution! Put the y value in terms of x on the right hand side of the second equation.
So, 4,194,304/x + 1 = 3,555,443.2/x + 2
=> (4,194,304 - 3,555,443.2) = x
which means: x = 638860.8
Put this into either and get y.