SOLUTION: can you help me solve this system of equations 2x+3y=2 2x^2+xy+y^2=8 thank you

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Question 224925: can you help me solve this system of equations
2x+3y=2
2x^2+xy+y^2=8
thank you
Answer by Edwin McCravy(6938) About Me  (Show Source):
You can put this solution on YOUR website!
system%282x%2B3y=2%2C%0D%0A2x%5E2%2Bxy%2By%5E2=8%29


The graph of this is a line intersecting a slanted ellipse.
You don't have to draw the graph, but I thought I'd graph
it for you so you'd know to expect two solutions:

drawing%28300%2C300%2C-6%2C6%2C-6%2C6%2C+%0D%0Agraph%28300%2C300%2C-6%2C6%2C-6%2C6%2C+%28-x%2Bsqrt%2832-7x%5E2%29%29%2F2%29%2C%0D%0Agraph%28300%2C300%2C-6%2C6%2C-6%2C6%2C+%28-x-sqrt%2832-7x%5E2%29%29%2F2%29%2C%0D%0Agraph%28300%2C300%2C-6%2C6%2C-6%2C6%2C+%282-2x%29%2F3%29%0D%0A%29

It looks as though one solution (where the line intersects
the ellipse is about (-2,2) and the other one looks about 
(2.1,-.8) or so.

Solve the first equation for x

2x%2B3y=2
2x=2-3y
x=%282-3y%29%2F2

Substitute in the second equation

2x%5E2%2Bxy%2By%5E2=8%29
2%28%282-3y%29%2F2%29%5E2%2B%28%282-3y%29%2F2%29y%2By%5E2=8%29

2%28%282-3y%29%5E2%2F2%5E2%29%2B%28%282-3y%29%2F2%29y%2By%5E2=8%29

2%28%282-3y%29%5E2%2F4%29%2B%28%282-3y%29%2F2%29y%2By%5E2=8%29

cross%282%29%28%282-3y%29%5E2%2Fcross%284%29%5B2%5D%29%2B%28%282-3y%29%2F2%29y%2By%5E2=8%29

%28%282-3y%29%5E2%2F2%29%2B%28%282-3y%29%2F2%29y%2By%5E2=8%29

%282-3y%29%5E2%2F2%2B%282y-3y%5E2%29%2F2%2By%5E2=8%29

Multiply through by 2 to clear of fractions:

%282-3y%29%5E2%2B%282y-3y%5E2%29%2B2y%5E2=16%29

%282-3y%29%5E2%2B2y-3y%5E2%2B2y%5E2=16%29

%282-3y%29%5E2%2B2y-y%5E2=16%29

%284-12y%2B9y%5E2%29%2B2y-y%5E2=16

4-12y%2B9y%5E2%2B2y-y%5E2=16

8y%5E2-10y%2B4=16

8y%5E2-10y-12=0

4y%5E2-5y-6=0

%28y-2%29%284y%2B3%29=0

y-2=0                  4y%2B3=0
y=2                    4y=-3
                         y=-3%2F4

x=%282-3y%29%2F2               x=%282-3y%29%2F2
x=%282-3%282%29%29%2F2             x=%282-3%28-3%2F4%29%29%2F2
x=%282-6%29%2F2                x=%282%2B9%2F4%29%2F2
x=%28-4%29%2F2                 x=%288%2F4%2B9%2F4%29%2F2
x=-2                  x=%2817%2F4%29%2F2
                         x=17%2F4%27%F7%272 
                         x=17%2F4%27%D7%271%2F2
                         x=17%2F8 

So the two solutions are:

%27%28x%2Cy%29%27=%27%28-2%2C2%29%27 and %27%28x%2Cy%29%27=%27%28%2717%2F8%27%2C%27-3%2F4%27%29%27

Edwin