Question 327883
<pre><b>
{{{system(25x^2+9y^2=225,
x-y=0)}}}

I'll do it first graphically and then I'll do it algebraically. 


The first equation has x intercepts (-3,0), (3,0). 
It also has y-intercepts (0,5), (0,-5), so it is this ellipse.

{{{drawing(400,400, -6,6,-6,6,
graph(400,400, -6,6,-6,6), arc(0,0,6,10)


  )}}} 

The second equation is of this line:

{{{drawing(400,400, -6,6,-6,6,
graph(400,400, -6,6,-6,6),

line(10,10, -10,-10) 

  )}}} 

Putting them on the same set of axes:

{{{drawing(400,400, -6,6,-6,6,
graph(400,400, -6,6,-6,6), arc(0,0,6,10),

line(10,10, -10,-10) 

  )}}} 

If we draw lines to the axes from the points of intersections

{{{drawing(400,400, -6,6,-6,6,
graph(400,400, -6,6,-6,6), arc(0,0,6,10),

line(10,10, -10,-10),

green(line(0,15/sqrt(34), 15/sqrt(34),15/sqrt(34))),

green(line(15/sqrt(34), 0,15/sqrt(34),15/sqrt(34))),

green(line(0,-15/sqrt(34), -15/sqrt(34),-15/sqrt(34))),

green(line(-15/sqrt(34), 0,-15/sqrt(34),-15/sqrt(34)))


 
  )}}}

It appears that the two points of intersection are (2.6,2.6)

and (-2.6,-2.6)

Now we'll do it algebraically to find the points exactly:

{{{system(25x^2+9y^2=225,
x-y=0)}}}

Use substitution.  Solve the second equation for y

{{{y=x}}}

So we substitute x for y is the first equation:

{{{25x^2+9y^2=225}}}

{{{25x^2+9x^2=225}}}

{{{34x^2=225}}}

{{{x^2=225/34}}}

{{{x = "" +- sqrt(225/34)}}}

{{{x = "" +- sqrt(225)/sqrt(34)}}}

{{{x = "" +- 15/sqrt(34)}}}

And since {{{y=x}}}

The points of intersection (exact values) are:

(x,y) = ({{{15/sqrt(34)}}}, {{{15/sqrt(34)}}})

(x,y) = ({{{-15/sqrt(34)}}}, {{{-15/sqrt(34)}}}) 

The decimal approximations are:

(x,y) = (2.572478777, 2.572478777)

(x,y) = (-2.572478777, -2.572478777)

Edwin</pre>