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Sum
Solve : `2/3"x" + 8 < 12; "x" ∈ "W"`
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Solution
`2/3"x" + 8 < 12`
⇒ `2/3"x" + 8 - 8 < 12 - 8`
⇒ `2/3"x" < 4`
⇒ `2/3"x" xx 3/2 < 4 xx 3/2` (Multiplying by `3/2`)
⇒ x < 6
∴ Required answer = {0, 1, 2, 3, 4, 5}
Concept: Replacement Set and Solution Set
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