SOLUTION: solve the following system of equations: x^2 + 4y^2=4 (x-1)^2+y^2=1

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Question 203724: solve the following system of equations:
x^2 + 4y^2=4
(x-1)^2+y^2=1
Found 2 solutions by Alan3354, jsmallt9:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
x^2 + 4y^2=4 --> 4y^2 = 4 - x^2
(x-1)^2+y^2=1 --> 4x^2 - 8x + 4 + 4y^2 = 4
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Sub for 4y^2 into 2nd eqn
4x^2 - 8x + 4 + (4-x^2) = 4
3x^2 - 8x + 4 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B-8x%2B4+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-8%29%5E2-4%2A3%2A4=16.

Discriminant d=16 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--8%2B-sqrt%28+16+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-8%29%2Bsqrt%28+16+%29%29%2F2%5C3+=+2
x%5B2%5D+=+%28-%28-8%29-sqrt%28+16+%29%29%2F2%5C3+=+0.666666666666667

Quadratic expression 3x%5E2%2B-8x%2B4 can be factored:
3x%5E2%2B-8x%2B4+=+%28x-2%29%2A%28x-0.666666666666667%29
Again, the answer is: 2, 0.666666666666667. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B-8%2Ax%2B4+%29

x = 2
x = 2/3
----------
at x=2: y = 0
--> (2,0) One solution
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at x = 2/3: y = +/- sqrt(8)/3
--> (2/3,y) 2nd point/solution
--> (2/3,-y) 3rd point/solution
----------------------------

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+%2B+4y%5E2=4
%28x-1%29%5E2%2By%5E2=1
Normally one use the substituion method to solve this type of system. You would start by solving for one of the variables in one of the equations and proceed from there. And this system can be done that way, too. But this system, since Y appears only as a squared term, can be solved in other ways:We can solve for y^2 instead of y and substitute for y^2.
  • We can use the Elimination (aka Linear Combination or Addition) Method.

    • I will use the first one of these.

    Solve the second equation for y^2:
    Subtract (x-1)^2 from both sides giving:
    y%5E2=+1+-+%28x-1%29%5E2
    Substitute into the first equation for y^2:
    x%5E2+%2B+4%281+-+%28x-1%29%5E2%29+=+4
    Solve for x. Start by simplifying:
    x%5E2+%2B+4%281+-+%28x%5E2+-2x+%2B+1%29%29+=+4
    x%5E2+%2B+4%281+-+x%5E2+%2B+2x+-+1%29+=+4
    x%5E2+%2B+4+-+4x%5E2+%2B+8x+-+4%29+=+4
    -3x%5E2+%2B+8x+=+4
    Make one side 0:
    0+=+3x%5E2+-+8x+%2B+4+=+0
    Factor: (or use the quadratic formula if you prefer)
    %283x+-+2%29%28x+-+2%29+=+0
    For this product to be zero one of the factors must be zero:
    3x+-+2+=+0 or x+-+2+=+0
    x+=+2%2F3 or x+=+2
    Now we use these, one at a time, to find the y values which go with them:
    For x = 2/3:
    %282%2F3%29%5E2+%2B+4y%5E2+=+4
    4%2F9+%2B+4y%5E2+=+4
    4y%5E2+=+32%2F9
    y%5E2+=+8%2F9
    y+=+sqrt%288%29%2F3+=+%282sqrt%282%29%29%2F3 or y = -%282sqrt%282%29%29%2F3
    For x = 2:
    %282%29%5E2+%2B+4y%5E2+=+4
    4+%2B+4y%5E2+=+4
    4y%5E2+=+0
    y+=+0

    So we have three points of intersection:
    (2/3, (2sqrt(2))/3)
    (2/3, -(2sqrt(2))/3)
    (2, 0)