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Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
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Solve $3 x+8 > 2$ when (i) x is an integer (ii) x is a real number

$\begin{array}{1 1} (0,0) \\ (0,\infty ) \\ (1,1) \\ (-2, \infty )\end{array}$

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Toolbox:
  • Same Quantity can be added (a subtracted ) to (from ) both sides of the inequality with out changing the sign of the in equality.
  • Same positive quantities can be multiplied or divided to both side of the in equality with out changing the sign of the inequality.
  • If same negative quantity is multiplied or divided to both sides of the inequality is reversed i.e $ '>'$ sign changes to $'<' $ and $'<'$ changes $'>'$ .
The given inequality is $3x+8 > 2$
Subtracting 8 from both sides,
=> $3x+8 -8 > 2-8$
=> $ 3x> -6$
Dividing by positive number 3 on both sides,
$=> x \geq -2$
Step 2:
Integer greater than -2 are $-1,0,1,2.....$
The solution set is $\{ -1,0,1,2,.\}$
Step 3:
when x is a real number all real numbers greater than -2 satisfy the inequality .
The solution set is $x \in (-2, \infty)$
answered Jul 25, 2014 by meena.p
 
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