Question

Solve the following linear inequations in R 

3x + 9 ≥ –x + 19

Answer 1

Given,

3x + 9 ≥ –x + 19 

⇒ 3x + 9 – 9 ≥ –x + 19 – 9 

⇒ 3x ≥ –x + 10 

⇒ 3x + x ≥ –x + 10 + x 

⇒ 4x ≥ 10

⇒ \(\frac{4x}{4}\) ≥ \(\frac{10}{4}\) 

∴ x ≥ \(\frac{5}{2}\)

Thus,

The solution of the given inequation is [\(\frac{5}{2}\),∞).