Question 1127257
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            I'd like to start noticing that the given system IS NOT a linear: it is the system of NON-LINEAR equations.



<pre>
The standard method to solve such a system is to introduce new variables

u = {{{1/x}}},  v = {{{1/y}}}.


Then the system takes the form


6u + 1v = 1     (1)
9u - 2v = 5     (2)


This you can solve by the Substitution method, for example.


For it express v = 1 - 6u  from (1) and substitute into eq(2).  You will get


9u - 2*(1-6u) = 5,

9u - 2 + 12u = 5,

21u = 5 + 2 = 7,

u = {{{7/21}}} = {{{1/3}}}.


Then from eq(1),  v = 1 - 6u = {{{1 - 6*(1/3)}}} = 1 - 2 = -1.


Now,  x = = {{{1/u}}} = {{{1/((1/3))}}} = 3  and  y = {{{1/v}}} = {{{1/(-1)}}} = -1.     


<U>ANSWER</U>.  x = 3  and  y = -1.
</pre>

Solved.


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Again,  <U>the lesson to learn from the solution is THIS</U> :


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The standard method to solve such a system is to introduce new variables.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;In this way, you reduce the original system of non-linear equations to the standard system of linear equations,

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;for which you can use many different and well known standard methods.