Question 1132110
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<pre>
{{{2/x}}} + {{{5/y}}} = {{{5/6}}},         (1) 

x/2 + y/5 = 5.       (2)


This system of equations in non-linear.

To make the solution easier, I will introduce NEW VARIABLES  a = {{{2/x}}},  b = {{{5/y}}}.  Then the system takes the form


a + b = {{{5/6}}}          (3)

{{{1/a}}} + {{{1/b}}} = 5         (4)


To solve it, first simplify equation (4):


    {{{1/a}}} + {{{1/b}}} = 5  ====> {{{(a+b)/(ab)}}} = 5  ====>  replace a+b by {{{5/6}}}, based on (3)  ====>  {{{((5/6))/(ab)}}} = 5  ====>  ab = {{{1/6}}}.


Next, from (3), express b = {{{5/6}}} - a  and substitute it into equation  ab = {{{1/6}}}.  You will get


    {{{a*(5/6-a)}}} = {{{1/6}}},

    -6a^2 + 5a = 1

    6a^2 - 5a + 1 = 0.


Solve the last quadratic equation using the quadratic formula.

You will get two solutions:  a= {{{1/2}}}  and  a= {{{1/3}}}.


Thus, the system  (3)-(4) has two solutions:


    1)  a= {{{1/2}}},  b= {{{5/6}}} - {{{1/2}}} = {{{2/6}}} = {{{1/3}}},     

and

    2)  a = {{{1/3}}},  b= {{{5/6}}} - {{{1/3}}} = {{{5/6 - 2/6}}} = {{{3/6}}} = {{{1/2}}}.


Now you need to return from "a" and "b" to x and y, via the formulas  a = {{{2/x}}},  b = {{{5/y}}}.


By doing so, you get two solutions for the original system:


1)  x = {{{2/a}}} = {{{2/((1/2))}}} = 4;    y = {{{5/b}}} = {{{5/((1/3))}}} = 15.


2)  x = {{{2/a}}} = {{{2/((1/3))}}} = 6;    y = {{{5/b}}} = {{{5/((1/2))}}} = 10.


<U>Answer</U>.  The system has two solutions  1)  x= 4; y= 15    and   2)  x= 6,  y= 10.
</pre>

Solved.


You may check that the solution is correct by substituting the found values into the original equations.



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I am very glad that the tutor @MathLover1 placed her solution here.


Comparing these two, &nbsp;you can see how many tons of calculations I saved you from, &nbsp;using my substitutions &nbsp;! &nbsp;! &nbsp;! 



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To see other similar solved problems for systems of two non-linear equations in two unknowns, &nbsp;look into the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/coordinate/lessons/Solving-systems-of-non-linear-equations-by-reducing-to-linear-ones.lesson>Solving systems of non-linear equations by reducing to linear ones</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Matrices-and-determiminant/Solving-systems-of-non-linear-equations-in-two-unknowns-using-Cramer%27s-rule.lesson>Solving systems of non-linear equations in two unknowns using the Cramer's rule</A> 

in this site.