\( \dfrac{(2x-1)}{3}≥\dfrac{(3x-2)}{4}-\dfrac{(2-x)}{5}\)
= \( \dfrac{(2x-1)}{3}≥\dfrac{5(3x-2)-4(2-x)}{20}\)
= \( \dfrac{(2x-1)}{3}≥\dfrac{15x-10-8+4x}{20}\)
= \( \dfrac{(2x-1)}{3}≥\dfrac{19x-18}{20}\)
= \( 20(2x-1)≥3(19x-18)\)
= \(40x-20≥57x-54\)
= \(-20+54≥57x-40x\)
= \(34≥17x\)
= \(2≥x\)
Thus, all real numbers x, which are less than or equal to 2, are the solutions of the given inequality.
Hence, the solution set of the given inequality is (-∞, 2]
Answered by Abhisek | 1 year agoSolve each of the following in equations and represent the solution set on the number line \( \dfrac{5x}{4}-\dfrac{4x-1}{3}>1,\) where x ϵ R.
Solve each of the following in equations and represent the solution set on the number line.\( \dfrac{5x-8}{3}\geq \dfrac{4x-7}{2}\), where x ϵ R.
Solve each of the following in equations and represent the solution set on the number line. 3 – 2x ≥ 4x – 9, where x ϵ R.
Solve each of the following in equations and represent the solution set on the number line. 3x – 4 > x + 6, where x ϵ R.
Solve each of the following in equations and represent the solution set on the number line. 5x + 2 < 17, where
(i) x ϵ Z,
(ii) x ϵ R.