Question

solve the equations:
coshx - 3sinhy = 0
2sinhx + 6coshy = 5

Answer 1

not sure the best approach ..

Answer 2

cramers rule

Answer 3

not been a mention of that in this chapter of the book

Answer 4

@amistre cramer or write as exponentials? i am not sure

Answer 5

i tried as exponentials maybe i need to persevere with that

Answer 6

\[{{ax\ by\ |\ n}\choose{cx\ dy\ |\ m}}\]
\[y=\frac{am-cn}{ad-cb}\]
\[x=\frac{bm-dn}{-(ad-cb)}\]

Answer 7

is cramer post A level?

Answer 8

i got no idea what level cramer is :)

Answer 9

coshx - 3sinhy = 0
2sinhx + 6coshy = 5

\[\frac{hmmm}{6cosh(y)cosh(x)}\]
never tried it with these things tho

Answer 10

im sure there are identities involved but i aint got good experience with the hyperbolic stuff:

i recall that: cos^2 - sin^2 = 1

Answer 11

i suspect its not the technique here, as i've never hear of cramers mentioned

Answer 12

i've tried that route it didn't get me anywhere either

Answer 13

i think i'll go back to exponential method see if it will yield like that

Answer 14

the exp route is turning them to e^(...) right?

Answer 15

yep

Answer 16

hmmm, this might be useful for a chk then:

http://www.wolframalpha.com/input/?i=coshx+-+3sinhy+%3D+0+and+2sinhx+%2B+6coshy+%3D+5

Answer 17

thanks

Answer 18

i got that but it didn't help
would it help to write
\[y=\sinh^{-1}(\frac{\cosh(x)}{3})\]

Answer 19

might do i'll check it out

Answer 20

multiplied first one by 2, added to second and got
\[2e^x+6e^{-y}=5\] but i am not sure that helps either

Answer 21

\[cosh(x)=\frac{e^x+e^{-x}}{2};\ sinh(x)=\frac{e^x-e^{-x}}{2}\]
\begin{array}l
cosh(x) - 3sinh(y) = 0\\
2sinh(x) + 6cosh(y) = 5\end{array}

\begin{array}l
\frac{e^x+e^{-x}}{2} - 3\frac{e^y-e^{-y}}{2} = 0\\
2\frac{e^x-e^{-x}}{2} + 6\frac{e^y+e^{-y}}{2} = 5\end{array}

\begin{array}l
e^x+e^{-x} - 3e^y+3e^{-y} = 0\\
2e^x-2e^{-x} + 6e^y+6e^{-y} = 10\end{array}

\begin{array}l
-2e^x-2e^{-x} +6e^y-6e^{-y} = 0\\
2e^x-2e^{-x} + 6e^y+6e^{-y} = 10\end{array}

\[-4e^{-x} +12e^y=10\]

\[-2e^{-x} +6e^y=5\]

would it be simpler to try a substitution?

\[cosh(x)=3sinh(y)\]

Answer 22

so maybe
\[y=\frac{arcsinh(x)}{3}\]

Answer 23

then
\[2\cos(x)+6\cosh(\frac{arcsinh(x)}{3})=5\] but then i am stuck again

Answer 24

meant
\[2\sinh(x)+6\cosh(\frac{arcsinh(x)}{3})=5\]