Question #269044

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


Expert's answer

for x = y:

6y>11  ⟹  y>11/6  ⟹  y>16y>11\implies y>11/6 \implies y>1


for x > y:

x=y+k,k>0x=y+k,k>0

6x−k>11  ⟹  x>(11+k)/66x-k>11\implies x>(11+k)/6

6y+5k>116y+5k>11

if y≤1y\le 1 then k>1  ⟹  x>2k>1\implies x>2


for x < y:

x=y−k,k>0x=y-k,k>0

6x+k>116x+k>11

6y−5k>11  ⟹  y>11/6+5k/6>16y-5k>11\implies y>11/6+5k/6>1


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


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