Question 998683
.
{{{27x^3+y^3}}} = {{{35}}}; 

{{{xy}}} = {{{2}}}.
 
Find 3x+y.
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Express  y  from the second equation  y = {{{2/x}}}  and  substitute into the first one.  You will get


{{{27x^3 + (8/x^3)}}} = {{{35}}},


Multiply both sides by  {{{x^3}}}.  You will get


{{{27x^6 + 8 }}} = {{{35x^3}}},     or


{{{27x^6 - 35x^3 + 8}}} = {{{0}}}.


Introduce new variable  z = {{{x^3}}}.  You will get a quadratic equation


{{{27z^2 - 35z + 8}}} = {{{0}}}.


Now,  apply the quadratic formula to solve it.  You will get 


z = {{{(35 +- sqrt(35^2 - 4*35*8))/54}}} = {{{(35 +- 19)/54}}}.


So,  {{{z[1]}}} = 1     ----->  x = 1,  y = 2.


{{{z[2]}}} = {{{8/27}}}     ----->  x = {{{2/3}}},  y = {{{3}}}.


<U>Answer</U>. &nbsp;Two solutions: 1) &nbsp;x = 1, &nbsp;y = 2; &nbsp;&nbsp;and &nbsp;2) &nbsp;x = {{{2/3}}}, &nbsp;y = {{{3}}}.