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Solve a system equation

by naoufelabs
Tags: equation, solve
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naoufelabs
#1
Dec18-12, 03:05 PM
P: 17
Hi everybody,
Please I want to Solve the system:

3x-2y=19
y3-2x=19


x,y real number!

Thank you !
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mfb
#2
Dec18-12, 05:07 PM
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P: 11,917
I think clever guessing (and showing that there are not other solutions) is the quickest method. You can solve one of those equations for a single variable and put this into the other equation, but I don't think you cannot simply find solutions like that.
dextercioby
#3
Dec18-12, 05:35 PM
Sci Advisor
HW Helper
P: 11,927
It looks like 3,3 is the only solution.

mfb
#4
Dec19-12, 06:46 AM
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P: 11,917
Solve a system equation

There is another solution near y=20.8216, x=13.1370.
naoufelabs
#5
Dec19-12, 09:02 AM
P: 17
Please I want the steps to find it, because I'm stumbled in the last steps as follow:
x*Ln(3)-y*Ln(2)=Ln(19)
3*Ln(y)-x*Ln(2)=Ln(19)

x*Ln(3)-y*Ln(2) - 3*Ln(y)-x*Ln(2) = 0
x*Ln(3)+x*Ln(2) = y*Ln(2)+3*Ln(y) {each one equals Ln(19)}
x*Ln(3)+x*Ln(2) = y*Ln(2)+3*Ln(y) = Ln(19)
x*Ln(6) = y*Ln(2)+3*Ln(y) = Ln(19)

x = ln(19)/ln(6)
.....

Here I'm stumbled.

Thank you for your response.
mfb
#6
Dec19-12, 09:12 AM
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P: 11,917
There is no useful way to simplify ln(19)/ln(6). It is just some number - and it is wrong.

The first step does not work. ln(a-b) is not the same as ln(a) - ln(b).
The third step looks wrong, too (where you say "{each one equals Ln(19)}").
Mark44
#7
Dec19-12, 11:20 AM
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P: 21,304
Quote Quote by naoufelabs View Post

3x-2y=19
y3-2x=19

x,y real number!
The first equation below is wrong, so everything below it is also invalid.

You apparently took the natural log of both sides, like so:
ln(3x - 2y) = ln(19)

Then you "distributed" the ln operation like this:
ln(3x) - ln(2y) = ln(19)
This is the step that is incorrect. There is no property of logs in which ln(A + B) = ln(A) + ln(B).

Quote Quote by naoufelabs View Post
Please I want the steps to find it, because I'm stumbled in the last steps as follow:
x*Ln(3)-y*Ln(2)=Ln(19)
3*Ln(y)-x*Ln(2)=Ln(19)

x*Ln(3)-y*Ln(2) - 3*Ln(y)-x*Ln(2) = 0
x*Ln(3)+x*Ln(2) = y*Ln(2)+3*Ln(y) {each one equals Ln(19)}
x*Ln(3)+x*Ln(2) = y*Ln(2)+3*Ln(y) = Ln(19)
x*Ln(6) = y*Ln(2)+3*Ln(y) = Ln(19)

x = ln(19)/ln(6)
.....

Here I'm stumbled.

Thank you for your response.
naoufelabs
#8
Dec19-12, 02:41 PM
P: 17
Thank you all.


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