Question

Find all ordered pairs `(x,y)` such that `x-xy^3 =7` and `xy^2 -xy =3`?

Answer 1

`(27/4, -1/3)" "` and `" "(1/4, -3)`

Explanation:

Given:

`{ (x-xy^3=7), (xy^2-xy=3) :}`

We can rewrite these two equations as:

`{ (x(1-y^3) = 7), (x(y^2-y) = 3) :}`

Hence:

`7/(1-y^3) = x = 3/(y^2-y)`

That is:

`-7/((y-1)(y^2+y+1)) = x = 3/(y(y-1))`

Multiply both ends by `(y-1)` to find:

`-7/(y^2+y+1) = 3/y`

Multiply both sides by `y(y^2+y+1)` to get:

`-7y = 3(y^2+y+1) = 3y^2+3y+3`

Add `7y` to both ends to get:

`0 = 3y^2+10y+3 = (3y+1)(y+3)`

Hence `y=-1/3` or `y=-3`

If `y=-1/3` then:

`x = 7/(1-(-1/3)^3) = 7/(1+1/27) = 7/(28/27) = 27/4`

If `y=-3` then:

`x = 3/((-3)^2-(-3)) = 3/(9+3) = 1/4`

So the only pairs which are solutions are:

`(27/4, -1/3)" "` and `" "(1/4, -3)`

Answer 2

See below.

Explanation:

`{(x-xy^3 = 7),(xy^2-xy=3):}` then

`(x-xy^3)/(xy^2-xy)=(1-y^3)/(y^2-xy) = ((1-y)(1+y+y^2))/((y-1)y) = 7/3` or

`-(1+y+y^2)=7/3 y` or solving for `y` we have `y = {-3,-1/3}`

but `x = 3/(y^2-y)` so we can compute the ordered pairs.

`{1/4,-3}` and `{27/4,-1/3}`