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# 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{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.



$\mathbf{SOLVED-EXAMPLE}$ 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: $$x \in [-1,10)$$



$\mathbf{PRACTICE}$ 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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Note by Priyansh Sangule
3 years, 3 months ago