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- Admin
- #1
- Feb 14, 2012
- 3,931
Solve the following system:
$x-y=1-z$
$3(x^2-y^2)=5(1-z^2)$
$7(x^3-y^3)=19(1-z^3)$
$x-y=1-z$
$3(x^2-y^2)=5(1-z^2)$
$7(x^3-y^3)=19(1-z^3)$
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System of Equations
- Thread starter anemone
- Start date
- Jan 26, 2012
- 183
My solution.
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.
$3(x-y)(x+y) = 5(1-z)(1+z)$ 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.