Let x and y be real numbers. Prove that, if 5x+y>11, then x>2 or y>1.

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for x = y:

$6y>11\implies y>11/6 \implies y>1$

for x > y:

$x=y+k,k>0$

$6x-k>11\implies x>(11+k)/6$

$6y+5k>11$

if $y\le 1$ then $k>1\implies x>2$

for x < y:

$x=y-k,k>0$

$6x+k>11$

$6y-5k>11\implies y>11/6+5k/6>1$

so, if 5x+y>11, then x>2 or y>1

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