Question 1064606
.
x^2/y +y^2/x =9 ........eq1

1/x +1/y= 3/4............eq2
solve the simultaneous equations
~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
{{{x^2/y +y^2/x}}} = 9       (1)

{{{1/x +1/y}}} = {{{3/4}}}.       (2)


In (1), multiply both sides by x*y. In 2), do the same. You will get an equivalent system (equivalent under the condition xy =/= 0):

{{{x^3 + y^3}}} = 9xy,       (3)

x + y = {{{(3/4)*xy}}}.     (4)


Now, {{{x^3 + y^3}}} = {{{(x + y)*(x^2 -xy + y^2)}}} = {{{(x+y)*((x^2 + 2xy + y^2) - 3xy)}}} = {{{(x+y)*((x+y)^2-3xy)}}}.


Next, replace here all instances of (x+y) by {{{(3/4)*xy}}}, according to (4), and you will get (3) in the form

{{{(3/4)*xy*((3/4)^2*(xy)^2 - 3xy)}}} = 9xy.


Cancel 3xy in both sides. You will get

{{{(9/16)*(xy)^2-3xy}}} = 12.


Multiply both sides by 16. You will get

{{{9(xy)^2 - 48xy - 12*16}}} = 0,

{{{3(xy)^2 - 16xy - 64}}} = 0.    (5)

Introduce new variable u = xy. Then (5) becomes

{{{3u^2 - 16u - 64}}} = 0.

Solve this quadratic equation using the quadratic formula. The roots are 

{{{u[1]}}} = 8  and/or  {{{u[2]}}} = {{{-8/3}}}.


Thus the system (3),(4) is reduced to two independent and much simpler systems:


1)  First system is 

    x + y = 6,
    xy    = 8


2) The second system is 

    x + y = -2,
    xy    = {{{-8/3}}}


The major reduction is done.

The first system has two solutions  (x,y) = (2,4)  and  (x,y) = (4,2).


The second system has two solutions  (x,y) = ({{{-1+sqrt(11/3)}}},{{{-1-sqrt(11/3)}}})  and  (x,y) = ({{{-1-sqrt(11/3)}}},{{{-1+sqrt(11/3)}}}).
</pre>

Carefully check my math.