How would I solve this exponential simultaneous equation?

Question

e^y + 5 - 9x = 0
y - ln(x + 4) = 2

Thanks.

Answer 1

e^y + 5 - 9x = 0
y - ln(x + 4) = 2

e^y + 5 - 9x = 0
y-2 = ln(x + 4) ===> e^(y-2) = x+4 ===> e^(-2)e^y=x+4

If you let z = e^y
then it is linear in z and x.

d:

Answer 2

e^y + 5 - 9x = 0
=>e^y = - 5 + 9x
=>lne^y = ln(- 5 + 9x )
=> ylne = ln(- 5 + 9x ) ......lne=1
=> y = ln(- 5 + 9x )
&
y - ln(x + 4) = 2
=>y = ln e^2 + ln(x + 4)
=>y = ln[e^2(x + 4)]


=> ln[e^2(x + 4)] =ln(- 5 + 9x )
=> e^2x +4e^2 = -5 + 9x
=> e^2x -9x = -5 - 4e^2
=> x ( e^2 -9) = -5 - 4e^2
=> x = ( -5 - 4e^2)/( e^2 -9)

use calculator & then find y

Answer 3

e^y = 9x - 5
y = ln (9x - 5)
ln (9x - 5) - ln (x + 4) = 2
ln [(9x - 5) / (x + 4) ] = 2
(9x - 5) / (x + 4) = e²
(9x - 5) = 7.39.(x + 4)
9x - 5 = 7.39 x + 29.6
1.61 x = 34.6
x = 21.5