Sep 10, 2013 Thread starter Admin #1 anemone MHB POTW Director Staff member Feb 14, 2012 3,963 Solve the following system: $x-y=1-z$ $3(x^2-y^2)=5(1-z^2)$ $7(x^3-y^3)=19(1-z^3)$
Sep 11, 2013 #2 Jester Well-known member MHB Math Helper Jan 26, 2012 183 My solution. Spoiler Clearly $x=y, z = 1$ is a solution so we exclude this. Factoring the second gives $3(x-y)(x+y) = 5(1-z)(1+z)$ which using the first gives $3(x+y) = 5(1+z)$. Thus using the first and this new equation we obtain $y = 4x-5$ and $z = 3x - 4\;\;\;(*)$. From the third factoring we obtain $7(x^2+xy+y^2) = 19(1+z+z^2)$. Substituting $(*)$ gives $-12(2x-3)(x-2) = 0$ giving $x = 2, 3/2$ and (*) gives the corresponding $y$ and $z$ values.
My solution. Spoiler Clearly $x=y, z = 1$ is a solution so we exclude this. Factoring the second gives $3(x-y)(x+y) = 5(1-z)(1+z)$ which using the first gives $3(x+y) = 5(1+z)$. Thus using the first and this new equation we obtain $y = 4x-5$ and $z = 3x - 4\;\;\;(*)$. From the third factoring we obtain $7(x^2+xy+y^2) = 19(1+z+z^2)$. Substituting $(*)$ gives $-12(2x-3)(x-2) = 0$ giving $x = 2, 3/2$ and (*) gives the corresponding $y$ and $z$ values.