- Thread starter
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- #1

- Feb 14, 2012

- 3,963

Solve the following system:

$x-y=1-z$

$3(x^2-y^2)=5(1-z^2)$

$7(x^3-y^3)=19(1-z^3)$

$x-y=1-z$

$3(x^2-y^2)=5(1-z^2)$

$7(x^3-y^3)=19(1-z^3)$

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,963

Solve the following system:

$x-y=1-z$

$3(x^2-y^2)=5(1-z^2)$

$7(x^3-y^3)=19(1-z^3)$

$x-y=1-z$

$3(x^2-y^2)=5(1-z^2)$

$7(x^3-y^3)=19(1-z^3)$

- Jan 26, 2012

- 183

$3(x-y)(x+y) = 5(1-z)(1+z)$ 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.