Solve the following inequalities for real x. 2x-3/4+ 9 ≥ 3 +4x/3 , x ∈ R. ​

Question

Solve the following inequalities for real x.

\(\frac{2x-3}{4}\) + 9 ≥ 3 + \(\frac{4x}{3}\) , x ∈ R.

Answer 1

We have,

\(\frac{2x-3}{4}\) + 9 ≥ 3 + \(\frac{4x}{3}\)

⇒ \(\frac{2x-3}{4}\) − \(\frac{4x}{3}\) ≥ 3 − 9

[Transposing \(\frac{4x}{3}\) to LHS and 9 to RHS]

⇒ \(\frac{3(2 − 3) − 4(4)}{12}\) ≥ −6

⇒ \(\frac{6 − 9 − 16}{12}\) ≥ −6

⇒ \(\frac{−9 − 10}{12}\)≥ −6

⇒ −9 − 10x ≥ −72

⇒ −10x ≥ −72 + 9

⇒ −10x ≥ −63

⇒ − \(\frac{−10x} {−10}\) ≤ \(\frac{−63}{−10}\)

⇒ x ≤ \(\frac{−63}{−10}\)

∴ x ∈ [ \(\frac{−63}{−10}\) , ∞)