# How solve system? x+y=7 lnx+lny=2ln2+ln3

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### 1 Answer

You should use the top equation to write `x` in terms of `y` , such that:

`x = 7 - y`

Replacing `7 - y` for `x` in the bottom equation yields:

`ln(7 - y) + lny = ln2^2 + ln 3`

`ln(7 - y) + lny = ln 4 + ln 3`

Converting the summation of logarithms into products yields:

`ln (y*(7 - y)) = ln (4*3)`

Equating the corresponding numbers yields:

`y*(7 - y) = 12 => -y^2 + 7y = 12 => y^2 - 7y + 12 = 0`

Using quadratic formula yields:

`y_(1,2) = (7+-sqrt(49 - 48))/2`

`y_(1,2) = (7+-1)/2 => y_1 = 4 ; y_2 = 3`

`x_1 = 7 - y_1 => x_1 = 7 - 4 => x_1 = 3`

`x_2 = 7 - y_2 => x_2 = 7 - 3 => x_2 = 4`

**Hence, evaluating the solutions to the simultaneous equations yields `x = 3, y = 4` and **`x = 4, y = 3.`