Find all ordered pairs #(x,y)# such that #x-xy^3 =7# and #xy^2 -xy =3#?
2 Answers
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
#-7/(y^2+y+1) = 3/y#
Multiply both sides by
#-7y = 3(y^2+y+1) = 3y^2+3y+3#
Add
#0 = 3y^2+10y+3 = (3y+1)(y+3)#
Hence
If
#x = 7/(1-(-1/3)^3) = 7/(1+1/27) = 7/(28/27) = 27/4#
If
#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)#
See below.
Explanation:
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