You need to convert the sum of logarithms into the logarithm of product, in the bottom equation, such that:

`ln x + ln y = ln (xy)`

Replacing `ln (xy)` for `ln x + ln y` yields:

`ln (xy) = ln 15 => xy = 15`

You should use the following special product in the top equation, such that:

`x^2 + y^2 = (x + y)^2 - 2xy`

Replacing `15` for `xy` yields:

`x^2 + y^2 = (x + y)^2 -2*15`

`x^2 + y^2 = (x + y)^2 -30`

Hence, you may re-write the system of equations, such that:

`{((x + y)^2 - 30 = 34),(xy = 15):}`

`{((x + y)^2 = 34 + 30),(xy = 15):}`

`{((x + y)^2 = 64),(xy = 15):}`

`{((x + y) = +-sqrt 64),(xy = 15):}`

`{((x + y) = +-8),(xy = 15):}`

You may use Lagrange's resolvents to create the following quadratic equation, such that:

`u^2 - (x+y)*u + xy = 0`

Considering `x + y = 8` , yields:

`u^2 - 8u + 15 = 0`

Using quadratic formula, yields:

`u_(1,2) = (8+-sqrt(64 - 60))/2 => u_(1,2) = (8+-2)/2`

`u_1 = 5; u_2 = 3 => {(x = 5),(y = 3),(x = 3),(y = 5):}`

Considering `x + y = -8` , yields:

`u^2 + 8u + 15 = 0`

Using quadratic formula, yields:

`u_(1,2) = (-8+-sqrt(64 - 60))/2 => u_(1,2) = (-8+-2)/2`

`u_1 = -3 ; u_2 = -5 => {(x = -5),(y = -3),(x = -3),(y = -5):}`

**Hence, evaluating the solutions to the given simultaneous equations, yields `{(x = +-5),(y = +-3),(x = +-3),(y = +-5):}` .**