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the given equation is |x-y| = |x-z| which when squared gives

=> x^2+y^2-2xy = x^2+z^2-2zy => y^2 - z^2 + 2zy - 2xy = 0 => (y-z)[(y+z)-2x] = 0 => (y+z)-2x = 0 From statement 2 it is 10 - 2x = 0 => x = 5

Let me know if mised something

humans - what I'm not sure is, in a condition where y = z so that y = z = 5, we have infinite solutions for the equation (x is any integer). How can it be B then? (your solution seems OK, so I can't make out what I'm missing) _________________

II. insufficient; if y and z are not 5, then x=5; however if y and z are equal to 5 then x can be any value. Thus, I and II combined allows you to solve for x.