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Simultaneous inequality? (Posted on 2023-07-15) Difficulty: 3 of 5
Solve following system equations:

3x+4y=26
√(x2+y2-4x+2y+5)+√(x2+y2-20x-10y+125)=10

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution | Comment 2 of 4 |
Let start with the second equation and complete the squares in the radicands. Then we have
sqrt[(x-2)^2 + (y+1)^2] + sqrt[(x-10)^2 + (y-5)^2] = 10

Now the two radicands describe the distance from (2,-1) to (x,y) and (10,5) to (x,y).  So what we actually have is an ellipse with foci of (2,-1) and (10,5) and the sum of distances equals 10.

But the distance from (2,-1) to (10,5) is also 10.  This means our ellipse is a degenerate ellipse consisting solely of the line segment from (2,-1) to (10,5).

The equation of the line passing through (2,-1) and (10,5) is 3x-4y=10.  Then the intersection of this line with 3x+4y=26 is (6,2).  This does lay on the segment spanning (2,-1) to (10,5), so it is a valid solution.

  Posted by Brian Smith on 2023-07-15 15:13:05
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