moongazer
  • moongazer
Find the value of x and y that will satisfy the system
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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moongazer
  • moongazer
e^x - y = -2 e^x + y = 1
anonymous
  • anonymous
is x-y and x+y both exponents
anonymous
  • anonymous
|dw:1349690570593:dw|

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anonymous
  • anonymous
Is that the question?
anonymous
  • anonymous
introduce In both sides
anonymous
  • anonymous
If that is the question then it has no real roots.
moongazer
  • moongazer
how do I solve it?
moongazer
  • moongazer
(e^x) - y = -2 (e^x) + y = 1
moongazer
  • moongazer
sorry for the late reply
anonymous
  • anonymous
Then Add equation one and two ........
anonymous
  • anonymous
i think no real solution... u might get5 imaginary....
hartnn
  • hartnn
still e^x = -1/2 has no real solution for x
anonymous
  • anonymous
Yes still no real solution
hartnn
  • hartnn
y comes out to be 3/2
hartnn
  • hartnn
u have choices/options ?
anonymous
  • anonymous
But that will not be true.
moongazer
  • moongazer
I don't have
anonymous
  • anonymous
imaginary solution is how?
anonymous
  • anonymous
|dw:1349691346340:dw|
anonymous
  • anonymous
|dw:1349691384296:dw|
moongazer
  • moongazer
that's also what I got
hartnn
  • hartnn
http://www.wolframalpha.com/input/?i=e%5Ex+%3D+-1%2F2
anonymous
  • anonymous
ok
moongazer
  • moongazer
Isn't it that e^(ln -1/2) =-1/2 ?
moongazer
  • moongazer
@hartnn @sauravshakya
hartnn
  • hartnn
yes, thats correct but y did u caculate that ?
moongazer
  • moongazer
nope
hartnn
  • hartnn
i meant yes, thats correct but why did u caculate that ?
anonymous
  • anonymous
e^(ln -1/2) =-1/2 this is hell wrong... !! the equation is not defined...
moongazer
  • moongazer
to find y
moongazer
  • moongazer
I got y=3/2
hartnn
  • hartnn
yes, there are no real solutions to this.....
anonymous
  • anonymous
impossible man..
moongazer
  • moongazer
What should I write on my paper?
hartnn
  • hartnn
i think u just write e^x=-1/2, y =3/2 and leave it
anonymous
  • anonymous
u can do the above thing or write no real solution...
hartnn
  • hartnn
*and write no real solution
moongazer
  • moongazer
I think I got the answer
jhonyy9
  • jhonyy9
ln of a negativ number - how it is possible ? --- little up check it please !!!
jhonyy9
  • jhonyy9
@hartnn ,@sauravshakya
moongazer
  • moongazer
I got x = ln -1 and y = 1
jhonyy9
  • jhonyy9
@sauravshakya
moongazer
  • moongazer
I think that is the answer
jhonyy9
  • jhonyy9
check where is defined the logarithm funvtion please
jhonyy9
  • jhonyy9
function --- sorry
moongazer
  • moongazer
I forgot to add e^x
jhonyy9
  • jhonyy9
e^x - y = -2 e^x + y = 1
hartnn
  • hartnn
ln(-1), 1 does not satisfy 2nd equation ln(-1) is imaginary = i\(\pi\)
jhonyy9
  • jhonyy9
ln(-1) how do you mean this ?
moongazer
  • moongazer
it should be ln -1/2 =x and 3/2=y
jhonyy9
  • jhonyy9
when by definition logb X so x need being allways positiv
hartnn
  • hartnn
\(e^{i\pi}=-1\) by euler's relation
jhonyy9
  • jhonyy9
e^x - y = -2 e^x + y = 1 so and for these equations ?
hartnn
  • hartnn
but if x is imaginary, y must be imaginary also, right ?
jhonyy9
  • jhonyy9
but in this case i think that is very simple us ln to both sides for every two equations and will get the right answers
moongazer
  • moongazer
when you substitute x with ln -1/2 it will be e^(ln -1/2) - y = -2 then -1/2-y=-2 y=3/2
moongazer
  • moongazer
I think my teacher doesn't need to know the decimal value of x.
hartnn
  • hartnn
ok, then write that....
moongazer
  • moongazer
ok, thanks for the answers :)
moongazer
  • moongazer
why is it ln (negative number) is undefined?
jhonyy9
  • jhonyy9
by domain definition of logarithms
jhonyy9
  • jhonyy9
you have learned it sure --- i think ,yes ?
moongazer
  • moongazer
ohh yes
moongazer
  • moongazer
because of the graph
jhonyy9
  • jhonyy9
good so than all is understood it right good luck bye
moongazer
  • moongazer
Thanks :) do you know how to decompose a function?
jhonyy9
  • jhonyy9
how do you think it ?
moongazer
  • moongazer
partial fractions the one with cases. I got an odd looking expression
moongazer
  • moongazer
i'll post it

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