Question 1181988
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<pre>

Regarding the first system of two equations,


express x = -2y  from the first equation and substitute to the second equation,
replacing x there.


You will get then


    (2y)^2 + y^2 = 5

     4y^2  + y^2 = 5

         5y^2    = 5

          y^2    = 1

          y      = {{{sqrt(1)}}} = +/- 1.


Thus you obtain two solutions:


    a)  for y = 1,   x = -2y = -2.


    b)  for y = -1,  x = -2y = 2.



<U>ANSWER</U>.  There are two solutions  (x,y) = (-2,1)  and  (x,y) = (2,-1).



These two solutions are two intersection points of the straight line  x + 2y = 0  and  the circle  x^2 + y^2 = 5. 
</pre>


Solved and carefully explained.


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See the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Systems-of-equations/Solving-the-system-of-alg-eqns-of-deg2-deg1.lesson>Solving systems of algebraic equations of degree 2 and degree 1</A> 

in this site.



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Solve the second system of equations similarly.



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