Solve the following linear inequations in R2(3-x) ≥ x/5 + 4

1.	Solve the following linear inequations in R2(3-x) ≥ x/5 + 4
Answer» Given, 2(3-x) ≥ x/5 + 4 ⇒ 6 - 2x ≥ \(\frac{x}{5}\)+ 4 ⇒ 6 - 2x ≥ \(\frac{x+20}{5}\) ⇒ (6 - 2x) x 5 ≥ (\(\frac{x+20}{5}\)) x 5 ⇒ 30 – 10x ≥ x + 20 ⇒ x + 20 ≤ 30 – 10x ⇒ x + 20 – 20 ≤ 30 – 10x – 20 ⇒ x ≤ 10 – 10x ⇒ x + 10x ≤ 10 – 10x + 10x ⇒ 11x ≤ 10 ⇒ \(\frac{11x}{11}\) ≤ \(\frac{10}{11}\) ∴ x ≤ \(\frac{10}{11}\) Thus, The solution of the given inequation is (-∞,\(\frac{10}{11}\)]. we have 2(3−x)≥5x+4x∈R 10(3−x)≥x+20 30−10x≥x+20 10≥11x x≤1110 so x∈(−∞,1110] Therefore x∈(−∞,1110]

Answer»

Given,

2(3-x) ≥ x/5 + 4

⇒ 6 - 2x ≥ \(\frac{x}{5}\)+ 4

⇒ 6 - 2x ≥ \(\frac{x+20}{5}\)

⇒ (6 - 2x) x 5 ≥ (\(\frac{x+20}{5}\)) x 5

⇒ 30 – 10x ≥ x + 20

⇒ x + 20 ≤ 30 – 10x

⇒ x + 20 – 20 ≤ 30 – 10x – 20

⇒ x ≤ 10 – 10x

⇒ x + 10x ≤ 10 – 10x + 10x

⇒ 11x ≤ 10

⇒ \(\frac{11x}{11}\) ≤ \(\frac{10}{11}\)

∴ x ≤ \(\frac{10}{11}\)

Thus,

The solution of the given inequation is (-∞,\(\frac{10}{11}\)].

we have