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Linear Inequalities
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Solve $-12 x > 30$ when (i) x is a natural number (ii) x is an integer
cbse
math
class11
ch6
linear-inequalities
exercise6-1
q2
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asked
Jul 25, 2014
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meena.p
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Jul 25, 2014
by
meena.p
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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 $'>'$ .
Step 1:
The given in equalities
$-12 x > 30$
Dividing both sides by negative number $-12$
=> $ \large\frac{-12 x}{-12} < \frac{30}{-12}$
=> $ x < \large\frac{-5}{2}$
Step 2:
There is no natural number less than $\large\frac{-5}{2}$
Step 3:
The integer less than $ \large\frac{-5}{2}$ are $.....-5,-4,-3$
The solution set is $ \{ ........,-5,-4,-3\}$
answered
Jul 25, 2014
by
meena.p
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