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

2 Answers
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)#

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}#