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.

I thank each one of you for reading this note. Comment below your views , answers and share interesting questions related to it.

Stay tuned for more notes and questions. You can even post your matter down here in the comments to make this topic more interesting. (I've reserved the advanced topics and properties for upcoming notes.)

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Complete set here : Inequalities

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Note by Priyansh Sangule
4 years ago

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