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Inequalities: Solving Linear Inequalities

Hello ! This is the second note in the series Inequalities.

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\[ \mathbf{LINEAR-INEQUALITIES} \]

  1. \( a \geq b \Rightarrow\) either \( a > b \) or \( a = b \)

  2. If \( a > b \) and \( b > c \Rightarrow \) \( a > c \) [ Transition property ]

  3. For \( a > b \Leftrightarrow a \pm c < b \pm c \)

  4. If \( a > b \), then: \[ \begin{align*} ac > bc & \text{ if} & c > 0 \\ ac < bc & \text{ if} & c < 0 \end{align*} \]

    In other words, the sign of inequality flips if we multiply a negative real number to both side. For example: \( 3 > 2 \Leftrightarrow -3 < -2 \) . Which is definitely true.

  5. If \( ax > b \), then: \[ \begin{align*} x > b/a & \text{ if} & a > 0 & \text{ [Sign remains unchanged] }\\ x < b/a & \text{ if} & a < 0 & \text{ [Sign changes] } \end{align*} \]

    In other words, the sign of inequality flips if we divide both sides by a negative real number.

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\[ \mathbf{SOLVED-EXAMPLE} \] 1. Find the range of values of \( x \) satisfying the inequality: \[ 3 \geq 1-2x > -19 \] Solution: \[ \begin{align*} -3 \leq 2x-1& < 19 & \text{ [Multiplying both sides by -1] } \\ -2 \leq 2x & < 20 & \\ -1 \leq x & < 10 & \text{ [Dividing both sides by 2] }
\end{align*} \] Thus the answer is: \( x \in [-1,10) \)

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\[ \mathbf{PRACTICE} \] 1. Find the range of values of \( y \) in interval form, satisfying the inequality: \[ -4 \leq \dfrac{y-1}{3} \leq a \] Where \( 'a' \) is number of irrational roots of: \[ \sqrt{x^2 + \sqrt{x^2 + 11}} + \sqrt{x^2 - \sqrt{x^2 + 11}} = 4 \]

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2. Algebraic mess

3. Frame your own questions and feel free to share it down below in the comments.

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Complete set here : Inequalities

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Note by Priyansh Sangule
2 years, 8 months ago

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