# Solve {(x^2 + 1 = 81 (y^2 + y)),( x^2 + x = 9 (y^3 + 1)):} ?

Sep 15, 2016

$y = \left\{0.0849261 , 0.146206 , 9.79971 , 9.99432\right\}$

and correspondingly for $x$

$x = \left\{2.54229 , - 3.546 , - 92.5827 , 94.3364\right\}$

#### Explanation:

$\left\{\begin{matrix}{x}^{2} + 1 = 81 \left({y}^{2} + y\right) \\ {x}^{2} + x = 9 \left({y}^{3} + 1\right)\end{matrix}\right.$

Solving for $x$ both equations

$\left\{\begin{matrix}x = \pm \sqrt{- 1 + 81 y + 81 {y}^{2}} \\ x = \frac{1}{2} \left(- 1 \pm \sqrt{37 + 36 {y}^{3}}\right)\end{matrix}\right.$

after squaring and equating

$- \frac{21}{2} + 81 y + 81 {y}^{2} - 9 {y}^{3} + \frac{1}{2} \sqrt{37 + 36 {y}^{3}} = 0$

so we get at

$p \left(y\right) = {\left(- \frac{21}{2} + 81 y + 81 {y}^{2} - 9 {y}^{3}\right)}^{2} - {\left(\frac{1}{2} \sqrt{37 + 36 {y}^{3}}\right)}^{2}$ or after expanding

$p \left(y\right) = 101 - 1701 y + 4860 {y}^{2} + 13302 {y}^{3} + 5103 {y}^{4} - 1458 {y}^{5} + 81 {y}^{6}$. and for $x$
$q \left(x\right) = 19684 + 56862 x + 69987 {x}^{2} - 15309 {x}^{3} - 8745 {x}^{4} + {x}^{6}$

Solving iteratively and dividing by the found roots, we get at

$y = \left\{0.0849261 , 0.146206 , 9.79971 , 9.99432\right\}$

and solving correspondingly for $x$

$x = \left\{2.54229 , - 3.546 , - 92.5827 , 94.3364\right\}$

Attached a plot showing the intersections of both curves.

so, the real solution pairs are:

( (x= -92.5827, y= 9.79971), (x= -3.546, y= 0.146206), (x= 2.54229, y = 0.0849261), (x= 94.3364, y= 9.99432) )