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

2 Answers
George C.
Jul 26, 2017

(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)

Cesareo R.
Jul 26, 2017

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}

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