Question 1007316
<pre>
Instead of doing your problem for you, I'll
do one EXACTLY like it in every detail, step
by step.  You can then do yours EXACTLY the
same way, using this one as a guide.
</pre>
Given the equation of an ellipse {{{x^2 +4y^2=8}}} and the 
equation of a line {{{y-x=-3}}}...Find the solutions 
algebraically.
<pre>
{{{system( x^2 +4y^2=8,y-x=-3)}}}

Solve the second equation for y

{{{y-x=-3}}}
   {{{y=x-3}}}

Substitute for y in the first equation:

{{{x^2+4(x-3)^2=8}}}

{{{x^2+4(x-3)(x-3)=8}}}

{{{x^2+4(x^2-6x+9)=8}}}

{{{x^2-4x^2-24x+36=8}}}

{{{5x^2-24x+36=8}}}

{{{5x^2-24x+28=0}}}

Factor:

{{{(5x-14)(x-2)=0}}}

5x-14=0;  x-2=0
 5x=14;    x=2
  x=14/5;

Substitute each in    

{{{y=x-3}}}

Substitute x=14/5

{{{y=14/5-3}}}
{{{y=14/5-15/5}}}
{{{y=-1/5}}}

So one point is (x,y) = (14/5,1/5)

Substitute x=2

{{{y=2-3}}}
{{{y=-1}}}

So the other point is (x,y) = (2,-1)

Now do yours exactly the same way
step-by-step.

Edwin</pre>