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

- moongazer

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

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- moongazer

e^x - y = -2
e^x + y = 1

- anonymous

is x-y and x+y both exponents

- anonymous

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## More answers

- anonymous

Is that the question?

- anonymous

introduce In both sides

- anonymous

If that is the question then it has no real roots.

- moongazer

how do I solve it?

- moongazer

(e^x) - y = -2
(e^x) + y = 1

- moongazer

sorry for the late reply

- anonymous

Then Add equation one and two ........

- anonymous

i think no real solution...
u might get5 imaginary....

- hartnn

still e^x = -1/2 has no real solution for x

- anonymous

Yes still no real solution

- hartnn

y comes out to be 3/2

- hartnn

u have choices/options ?

- anonymous

But that will not be true.

- moongazer

I don't have

- anonymous

imaginary solution is how?

- anonymous

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- anonymous

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- moongazer

that's also what I got

- hartnn

http://www.wolframalpha.com/input/?i=e%5Ex+%3D+-1%2F2

- anonymous

ok

- moongazer

Isn't it that e^(ln -1/2) =-1/2 ?

- moongazer

@hartnn @sauravshakya

- hartnn

yes, thats correct but y did u caculate that ?

- moongazer

nope

- hartnn

i meant yes, thats correct but why did u caculate that ?

- anonymous

e^(ln -1/2) =-1/2 this is hell wrong... !!
the equation is not defined...

- moongazer

to find y

- moongazer

I got y=3/2

- hartnn

yes, there are no real solutions to this.....

- anonymous

impossible man..

- moongazer

What should I write on my paper?

- hartnn

i think u just write e^x=-1/2, y =3/2 and leave it

- anonymous

u can do the above thing or write no real solution...

- hartnn

*and write no real solution

- moongazer

I think I got the answer

- jhonyy9

ln of a negativ number - how it is possible ? --- little up check it please !!!

- jhonyy9

@hartnn ,@sauravshakya

- moongazer

I got x = ln -1
and y = 1

- jhonyy9

@sauravshakya

- moongazer

I think that is the answer

- jhonyy9

check where is defined the logarithm funvtion please

- jhonyy9

function --- sorry

- moongazer

I forgot to add e^x

- jhonyy9

e^x - y = -2
e^x + y = 1

- hartnn

ln(-1), 1 does not satisfy 2nd equation
ln(-1) is imaginary = i\(\pi\)

- jhonyy9

ln(-1) how do you mean this ?

- moongazer

it should be ln -1/2 =x
and 3/2=y

- jhonyy9

when by definition logb X so x need being allways positiv

- hartnn

\(e^{i\pi}=-1\)
by euler's relation

- jhonyy9

e^x - y = -2
e^x + y = 1
so and for these equations ?

- hartnn

but if x is imaginary, y must be imaginary also, right ?

- 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

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

I think my teacher doesn't need to know the decimal value of x.

- hartnn

ok, then write that....

- moongazer

ok, thanks for the answers :)

- moongazer

why is it ln (negative number) is undefined?

- jhonyy9

by domain definition of logarithms

- jhonyy9

you have learned it sure --- i think ,yes ?

- moongazer

ohh yes

- moongazer

because of the graph

- jhonyy9

good so than all is understood it right
good luck
bye

- moongazer

Thanks :)
do you know how to decompose a function?

- jhonyy9

how do you think it ?

- moongazer

partial fractions the one with cases.
I got an odd looking expression

- moongazer

i'll post it

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