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Find the value of x and y that will satisfy the system

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e^x - y = -2 e^x + y = 1
is x-y and x+y both exponents

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Other answers:

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

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