e^2y - x + 2 = 0
ln(x+3) - 2y - 1 = 0
need help on how to start solving this! thanks
Hello, jaderberg!
This is not pretty . . .
. e^{2y} - x + 2 .= .0
ln(x + 3) - 2y - 1 .= .0
We have: .e^{2y} .= .x - 2 - - [1]
. . . . . . .ln(x + 3) .= .2y + 1 . [2]
Equation [1] becomes: -ln(x - 2) .= .2y
. .Subtract it from [2]: .ln(x + 3) .= .2y + 1
We have: .ln(x + 3) - ln(x - 2) .= .1 . → . ln(x + 3) .= .1 + ln(x - 2)
. . ln(x + 3) .= .ln(e) + ln(x - 2) . → . ln(x + 3) .= .ln[e(x - 2)]
Un-log both sides: .x + 3 .= .ex - 2e . → . ex - x .= .2e + 3
Factor: .x(e - 1) .= .2e + 3
. . Hence: . x .= .(2e + 3)/(e - 1)
Substitute into [1]: .e^{2y} .= .(2e + 3)/(e - 1) - 2 .= .5/(e - 1)
. . Hence: .2y .= .ln[5/(e - 1)] . → . y .= .½·ln[5/(e - 1)]