Question

How do you solve the system `(x+y)^2+3(x-y) = 30` and `xy+3(x-y) = 11` ?

Answer 1

`(x, y)` is one of:

`(1+sqrt(6), -1+sqrt(6))`

`(1-sqrt(6), -1-sqrt(6))`

`(5, -2)`

`(2, -5)`

Explanation:

Notice that:

`(x+y)^2-4xy = (x-y)^2`

So we can get a quadratic in `(x-y)` by subtracting `4` times the second equation from the first...

`-14 = 30 - 4*11`

`color(white)(-14) = ((x+y)^2+3(x-y))-4(xy + 3(x-y))`

`color(white)(-14) = ((x+y)^2-4xy)+(3-12)(x-y)`

`color(white)(-14) = (x-y)^2-9(x-y)`

Add `14` to both ends to get:

`0 = (x-y)^2-9(x-y)+14`

`color(white)(0) = ((x-y)-2)((x-y)-7)`

So `x-y = 2` or `x-y = 7`

`color(white)()`
Case `x-y = 2`

From the first given equation, we have:

`30 = (x+y)^2+3(x-y)`

`color(white)(30) = (2y+2)^2+6 = 4(y+1)^2 + 6`

Subtract `6` from both ends and transpose to get:

`4(y+1)^2 = 24`

Hence:

`(y+1)^2 = 6`

So:

`y + 1 = +-sqrt(6)`

So

`y = -1+-sqrt(6)`

with corresponding values of `x` given by `x = y+2`

So solutions:

`(x, y) = (1+sqrt(6), -1+sqrt(6))`

`(x, y) = (1-sqrt(6), -1-sqrt(6))`

`color(white)()`
Case `x-y = 7`

From the first given equation, we have:

`30 = (x+y)^2+3(x-y)`

`color(white)(30) = (2y+7)^2+21`

Subtract `21` from both ends and transpose to get:

`(2y+7)^2 = 9`

Hence:

`2y+7 = +-sqrt(9) = +-3`

So:

`2y = -7+3 = -4`

or:

`2y = -7-3 = -10`

Hence `y = -2` or `y = -5`

Then we have corresponding values for `x` using `x = y+7`

Hence solutions:

`(x, y) = (5, -2)`

`(x, y) = (2, -5)`

graph{((x+y)^2+3(x-y)-30)(xy+3(x-y)-11) = 0 [-7.22, 12.78, -6.36, 3.64]}