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1a) 2x+1=7+1=8, so x=7/2; 2x+1=7-1=6, so x=5/2. So 5/2<x<7/2, x=(2.5,3.5) or x=3±0.5. 0.5 is half a unit.

b) the clue is to halve the difference specified for the limit, so x=3±0.3, then 2x+1=7±0.6 or x=(2.7,3.3) or 2.7<x<3.3.

c) x=3±0.02, x=(2.98,3.02) or 2.98<x<3.02 d) x=3±ɛ/2, x=(3-ɛ/2,3+ɛ/2) or 3-ɛ/2<x<3+ɛ/2.

2a) x=1±⅓, x=(⅔,1⅓) or ⅔<x<1⅓ b) x=1±0.2,... c) x=1±0.03,... d) x=1±ɛ/3,...

where ... simply means the alternative formats:  () or inequality. To be within the limits the ± format may need to be omitted because it implies less than or equal instead of less than.

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