Question 985922
.
{{{system(y=x^2-10,
9x^2-y^2=90)}}}.


Substitute  {{{y}}} = {{{x^2}}} - {{{10}}}  from the first equation into the second equation.  You will get


{{{9*x^2}}} - {{{(x^2-10)^2}}} = {{{90}}}.


Solve it:


{{{9x^2}}} - ({{{x^4 - 20*x^2 + 100}}}) = {{{90}}},


{{{9x^2}}} - {{{x^4}}} + {{{20x^2}}} - {{{100}}} = {{{90}}},


{{{x^4}}} - {{{29x^2}}} + {{{190}}} = {{{0}}}.


It is a <B>bi-quadratic</B> equation. &nbsp;&nbsp;&nbsp;&nbsp;Don't be afraid ! &nbsp;&nbsp;&nbsp;&nbsp;Don't panic !


The Math gives you an opportunity to demonstrate your skills ! :)


Introduce new variable &nbsp;{{{z}}} = {{{x^2}}}. 


Then you will get quadratic equation


{{{z^2}}} - {{{29z}}} + {{{190}}} = {{{0}}}.


Its roots are &nbsp;(use the quadratic formula)


{{{z[1]}}} = 19, {{{z[2]}}} = 10. 


Hence, &nbsp;{{{x}}} = +/-{{{sqrt(19)}}}, +/-{{{sqrt(10)}}}.


Then from the first equation of the system


y = {{{x^2}}} - {{{10}}} = 19 - 10 = 9 &nbsp;&nbsp;OR&nbsp;&nbsp; y = 10 - 10 = 0.


<B>Answer</B>. &nbsp;The solutions are the pairs &nbsp;(x,y) = ({{{sqrt(19)}}}, {{{9}}}), &nbsp;({{{-sqrt(19)}}}, {{{9}}}), &nbsp;({{{sqrt(10)}}}, {{{0}}}) &nbsp;and &nbsp;({{{-sqrt(10)}}}, {{{0}}}).