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A real number line can be used to graphing linear inequality.The convention is that O ( hollow/open circle ) marks the end of range with a strict inequality ( i.e < or >) and the • ( closed circle) marks the end of the range involving an equality as well (i.e. $\geq$ or $\leq$).

The number 4 is encircled and the circle is open not closed to show that 4 is not included in the graph or you can say that 4 is not the value of x.

If 4 is not included i.e. x $\geq$ 4 then the there will be closed circle at 4 and the graph will look like this. |

As there is positive 9, so we will add negative(-9) on both sides

2x + 9 - 9 > 15 - 9

2x > 6

As there is a multiplication between 2 and x so to isolate x we will divide both side by 2

$\frac{2x}{2}$ > $\frac{6}{2}$

∴ x > 3

2) -3x - 18 $\geq$ - 15

As there is negative 18, so we will add positive 18 on both sides

-3x -18 + 18 $\geq$ - 15 + 18

-3x $\geq$ 3

As there is a multiplication between -3 and x so to isolate x we will divide both side by -3 and since we are dividing both sides by negative number

$\frac{-3x}{-3} \leq \frac{3}{-3}$

x $\leq$ -1

(I) In which of the following inequality, state whether there is open/closed circle.

(i) 2x > 4

(ii) 3x < 6

(iii) x $\geq$ -4

(iv) - 2x $\leq$ 12

(v)$\frac{x}{3}$ $\leq$ 4

(vi) $\frac{-3x}{7}$ $\geq$ 3

(vii) 2x $\leq$ -4

(viii) 4x > 16

II) Represent the following inequalities on real number lines :

(i) 2x - 2 < 4

(ii) 3x + 1 $\leq$ -5

(iii) -2 $\leq$ < 5

(iv) 8 $\geq$ - 4

(v) -2 < x < 3

(vi) 4x + 8> 12

(vii) -5x $\geq$ -5

(viii) 11x - 5 < 50

(ix) 12x $\leq$ 96

(x) $\frac{7x}{5}$ $\geq$ 42

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