Question 6845
{{{ 7x^2+11y^2=28 }}}    ---->(1)
{{{ x^2+y^2=4 }}}       --->(2)  
This time we have an ellipse and a hyperbola.  Neither one are in standard form however

if we just add the two equations we will eliminate the y’s from the system 
for that multiply equation(2) by -11 we get

{{{-11x^2 -11y^2 =-44}}}  --->(3)
Now adding equation (1) and equation(3) we get

{{{ 7x^2  +11y^2  =  28}}}
{{{-11x^2 -11y^2  = -44}}}
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{{{-4x^2    +0    = -16}}}

solve for x.
{{{-4x^2 = -16}}}
{{{x^2 = 16/4 = 4}}}
{{{x^2=4}}}
{{{x= sqrt(4) =2}}}
x= +/-2
To determine the value(s) of the y’s we can substitute these into either of the equations.
x=2
      {{{7x^2  +11y^2  =  28}}}
      {{{7(2)^2 +11y^2 =28}}}
      {{{28 +11y^2 =28}}}
 
       {{{y^2 =0}}}
      {{{y=0}}}
x=-2
     {{{7(-2)^2 +11y^2 =28}}}

      {{{y=0}}}
Therefore the solutions are :(2,0) (-2,0)