Question 409847
Square the linear equation, giving {{{x^2 + 4xy + 4y^2 = 49}}}.  Subtract the equation {{{x^2 + 4y^2 = 25}}} from the preceding equation, to get 4xy = 24, or xy = 6, or y = 6/x.
Then
{{{x^2 + 4(6/x)^2 = 25}}}
<==> {{{x^2  + 144/x^2 = 25}}}
<==> {{{x^4  - 25x^2 + 144 = 0}}}
<==> {{{(x^2 - 16)(x^2 - 9) = 0}}}
==> x = 4, -4, 3, -3
The graph of  {{{x^2 + 4y^2 = 25}}} is an ellipse with vertices (-5,0) and (5,0); and co-vertices (0, 5/2) and (0, -5/2).  The graph of  x +2y = 7 is a line with slope -1/2 and y-intercept 7/2.  The two graphs intersect only at two points (with both x-coordinates positive).  Hence the intersection points are:
When x = 4, y = 6/4 = 3/2 ==> (4, 1.5).
When x = 3, y = 6/3 = 2 ==> (3,2).