Advertisements
Advertisements
Question
Solve : `2/3"x" + 8 < 12; "x" ∈ "W"`
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}
APPEARS IN
RELATED QUESTIONS
If the replacement set is the set of natural numbers, solve: x + 1 ≤ 7
If the replacement set is the set of natural numbers, solve: 3x - 4 > 6
If the replacement set is the set of natural numbers, solve: 4x + 1 ≥ 17
If the replacement set = {- 6,. - 3, 0, 3, 6, 9}; find the truth set of the following : 3x + 7 < 1
Solve : 7 > 3x - 8 ; x ∈ N.
Solve the inequation 18 – 3 (2x – 5) > 12; x ∈ W.
Solve the inequation `("2x" + 1)/3 + 15 ≤ 17;` x ∈ W.
Solve the inequation -3 + x < 2, x ∈ N.
Solve: 4x - 5 > 10 - x, x ∈ {0, 1, 2, 3, 4, 5, 6, 7}.
Solve: `("2x" + 3)/5 > ("4x" - 1)/2`, x ∈ W.
