Inequalities: Solving Linear Inequalities

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



$\mathbf{LINEAR-INEQUALITIES}$

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{aligned} ac > bc & \text{ if} & c > 0 \\ ac < bc & \text{ if} & c < 0 \end{aligned}

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{aligned} x > b/a & \text{ if} & a > 0 & \text{ [Sign remains unchanged] }\\ x < b/a & \text{ if} & a < 0 & \text{ [Sign changes] } \end{aligned}

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



$\mathbf{SOLVED-EXAMPLE}$ 1. Find the range of values of $x$ satisfying the inequality: $3 \geq 1-2x > -19$ Solution: \begin{aligned} -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{aligned} Thus the answer is: $x \in [-1,10)$



$\mathbf{PRACTICE}$ 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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Complete set here : Inequalities

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Note by Priyansh Sangule
5 years, 2 months ago

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