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)]