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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, with some Wolfram help | Comment 1 of 4
I looked to see where the first equation would have (x,y) as integers:
it's true for (6,2).

Plugging those values into the second equation shows that it is a solution.
Lucky guess.
You can factor the second equation as:
√((x-2)^2 + (y+1)^2) + √((x-10)^2 + (y-5)^2)

Plugging in (6,2):
√(16 + 9) + √(16 + 9) = 5+5 = 10

So, (6,2) is a solution.

Now, with a little (or a lot of) help from Wolfram Alpha:
Putting the second equation (without =10) into Wolfram Alpha shows that the z value is always >= 10.
Setting it equal to 10 shows a real solution to be:
2 <= x <= 10 and y = (1/4)(3x - 10) or 3x-4y = 10

3x + 4y = 26
3x - 4y = 10
intersects at (6,2) so (I think) this is the only solution.

  Posted by Larry on 2023-07-15 14:04:48
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