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Solve the following linear inequations in R 3x + 9 ≥ –x + 19 |
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Answer» 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 [\(\frac{5}{2}\),∞). |
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