Question 430768
{{{3x^2+2y^2-29=0}}}
{{{2x^2-5y^2-13=0}}}
<pre>
Do it by the addition method.
Eliminate {{{y^2}}} by multiplying the first equation through by 5
and the second equation through by 2.


{{{red(5)3x^2+red(5)2y^2-red(5)29=red(5)0}}}
{{{red(2)2x^2-red(2)5y^2-red(2)13=red(5)0}}}


{{{15x^2+10y^2-145=0}}}
 {{{4x^2-10y^2-26=0}}}
{{{19x^2-171=0}}}
{{{x^2-9=0}}}
{{{x^2=9}}}
{{{x=""+- 3}}} 


Substitute ±3 for x in

{{{3x^2+2y^2-29=0}}}
{{{3(""+- 3)^2+2y^2-29=0}}}
{{{3(9)+2y^2-29=0}}}
{{{27+2y^2-29=0}}}
{{{2y^2-2=0}}}
{{{y^2-1=0}}}
{{{y^2=1}}}
{{{y=""+- 1}}}

So there are 4 solutions:

(3,1), (3,-1), (-3,1), and (-3,-1)

Here is the graphical solution:

{{{drawing(400,400,-8,8,-8,8, 
graph(400,400,-8,8,-8,8, sqrt( (2x^2-13)/5 )),
graph(400,400,-8,8,-8,8,-sqrt( (2x^2-13)/5 )),
arc(0,0,-2sqrt(29/3),2sqrt(29/2))
 )}}}

The black oval curve is the graph of the first equation.
The red curve (in two parts) is the graph of the second equation
Notice that they intersect at those 4 solution points.

Edwin</pre>