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Thus;

(i) to solve the inequality 2x + 4 > 6 means to find the variable x.

(ii) to solve the inequality 5 - 6y $\leq$ 3 means to find the value of y and so on.

The following are the rules must be followed for solving a given linear inequality.

4x + 6 > 7 ⇒ 4x > 7 - 6, 21 $\geq$ 2x + 8 ⇒ 21 - 8 $\geq$ 2x

7x + 4 $\leq$ 18 ⇒ 7x $\leq$ 18 - 4 and so on.

3x - 6 > 5 ⇒ 3x > 5 + 6, 42 $\geq$ 2x - 9 ⇒ 42 + 9 $\geq$ 2x

9x - 1 $\leq$ 20 ⇒ 9x $\leq$ 20 + 1 and so on.

x < y ⇒ px < py and $\left ( \frac{x}{p} \right )$ < $\left ( \frac{x}{p} \right )$

x > y ⇒ px > py and $\left ( \frac{x}{p} \right )$ > $\left ( \frac{x}{p} \right )$

x $\leq$ y ⇒ px $\leq$ py and $\left ( \frac{x}{p} \right )$ $\leq$ $\left ( \frac{x}{p} \right )$

x $\geq$ y ⇒ px $\geq$ py and $\left ( \frac{x}{p} \right )$ $\geq$ $\left ( \frac{x}{p} \right )$

For example x $\leq$ 5 ⇒ 3x $\leq$ 3 x 5 and x $\geq$ 5 ⇒ 3x $\geq$ 3 x 5

Case II : If p is negative,

x < y ⇒ -px > -py and $\left ( \frac{-x}{p} \right )$ > $\left ( \frac{-x}{p} \right )$

x > y ⇒ -px < -py and $\left ( \frac{-x}{p} \right )$ > $\left ( \frac{-x}{p} \right )$

x $\leq$ y ⇒ -px $\geq$ -py and $\left ( \frac{-x}{p} \right )$ $\geq$ $\left ( \frac{-x}{p} \right )$

x $\geq$ y ⇒ -px $\leq$ -py and $\left ( \frac{-x}{p} \right )$ $\leq$ $\left ( \frac{-x}{p} \right )$

(i) x > y $\Leftrightarrow \frac{1}{x}$ < $\frac{1}{y}$ (ii) $x \geq y \Leftrightarrow \frac{1}{x} \leq \frac{1}{y}$

(iii) $x \leq y \Leftrightarrow \frac{1}{x} \geq \frac{1}{y}$ and so on.

11th grade math

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