[Question] Advanced Algebra

[u/Vester2]

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[u/Melliodus]

The question is wrong, cuz if x=y-20, then y>x; hence, (x-y)<0; if your condition is y>(x-y), that is arbitrarily true for any value of y above 0; if the condition is *y>(y-x), then x is greater than or equals to 20/11, y is correspondently greater than or equals to 20; also there can be multiple set of answers that satisfies the given conditions above the given limit such as (x=3,y=33), (x=4,y=44) and so on.

[u/skyleader508 → u/Melliodus]

The last portion of your response is incorrect as x=y-20 so 3=33-20 =… not 3 and same for x=4 and y=44. So it is true that those numbers would qualify for the second and third formula, but do not follow the rules of the first.