Question 1066254
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{{{9x^2+25y^2}}} = 40     (1)
{{{3x+5y}}} = 6         (2)


Express 3x = 6 - 5y from (2) and then substitute it into (1), replacing {{{9x^2}}} = {{{(3x)^2}}}  (!!).  You will get

{{{(6-5y)^2 + 25y^2}}} = 40,

{{{36 - 60y + 25y^2 + 25y^2}}} = 40,

{{{50y^2 - 60y - 4}}} = 0.

{{{y[1,2]}}} = {{{(60 +- sqrt(3600 + 4*4*50))/100}}} = {{{(60 +- 10*sqrt(44))/100}}} = {{{0.6 +- 0.2*sqrt(11)}}}.


Thus,  one solution is {{{y[1]}}} = {{{0.6 + 0.2*sqrt(11)}}}, {{{x[1]}}} = {{{(6-5y[1])/3}}} = {{{(3 - sqrt(11))/3}}}.

The second solution is {{{y[2]}}} = {{{0.6 - 0.2*sqrt(11)}}}, {{{x[2]}}} = {{{(6-5y[2])/3}}} = {{{(3+sqrt(11))/3}}}.
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