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

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\[ \mathbf{LINEAR-INEQUALITIES} \]

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

If \( a > b \) and \( b > c \Rightarrow \) \( a > c \)

**[**Transition property**]**For \( a > b \Leftrightarrow a \pm c < b \pm c \)

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.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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