Advertisements
Advertisements
प्रश्न
Solve:
`x/2 + 5 <= x/3 + 6`, where x is a positive odd integer
उत्तर
`x/2 + 5 <= x/3 + 6`
`x/2 - x/3 <= 6 - 5`
`x/6 <= 1`
`x <= 6`
Since, x is a positive odd integer
∴ Solution set = {1, 3, 5}
APPEARS IN
संबंधित प्रश्न
Solve the following inequation, write the solution set and represent it on the number line `-x/3 <= x/2 - 1 1/3 < 1/6, x ∈ R`
Represent the following inequalities on real number line:
– 2 ≤ x < 5
Represent the solution of the following inequalities on the real number line:
7 – x ≤ 2 – 6x
Solve the following in equation write the solution set and represent it on the number line:
`-x/3 <= x/2 -1 1/3 < 1/6, x ∈ R`
Find the values of x, which satisfy the inequation
`-2 5/6 < 1/2 - (2x)/3 <= 2`, x ∈ W
Graph the solution set on the number line.
Solve the following linear in-equation and graph the solution set on a real number line :
3x - 9 ≤ 4x - 7 < 2x + 5, x ∈ R
Solve the following inequation, write the solution set and represent it on the number line.
`-3 (x - 7) ≥ 15 - 7x > (x + 1)/3, x ∈ R`
Graph the solution set for each inequality:
5 ≤ x < 10
Solve the following inequalities and represent the solution set on a number line:
`0 ≤ (3 - 2x)/(4) ≤ 1`
Given that x ∈ I, solve the inequation and graph the solution on the number line: `3 ≥ (x - 4)/(2) + x/(3) ≥ 2`
