Q32 of 32 Page 107

Fill in the blanks of the following:

(i) If – 4x ≥ 12, then x ... – 3.


(ii) If x ≤ –3, then x…..4.


(iii) If > 0, then x …. –2.


(iv) If x > –5, then 4x ... –20.


(v) If x > y and z < 0, then – xz ... – yz.


(vi) If p > 0 and q < 0, then p – q ... p.


(vii) If |x+ 2|> 5, then x ... – 7 or x ... 3.


(viii) If – 2x + 1 ≥ 9, then x ... – 4.

(i) ≤


-4x ≥ 12


Dividing by 4, we get


⇒ -x ≥ 3


Now, taking negative both sides inverts the inequality sign


⇒ x ≤ 3


(ii) ≤



Multiplying by 4, we get


⇒ -3x ≥ -12


Now, taking negative both sides invert the inequality sign


⇒ 3x ≤ 12


⇒ x ≤ 4


(iii) >



Now, for above to be greater than 0,


x + 2 > 0


⇒ x > -2


(iv) >


x > -5


Multiplying by 4 both sides, we get


⇒ 4x > -20


(v) >


x > y


Since, z < 0, i.e. z is negative and multiplying by negative inverts the inequality sign


⇒ xz < yz


Now, taking negative both sides invert the inequality sign


⇒ -xz > -yz


(vi) >


q < 0


Now, taking negative both sides invert the inequality sign


⇒ -q > 0


Adding p both side, we get


⇒ p – q > p [since, p>0 inequality sign retains the same]


(vii) <, >


Given,


|x + 2| > 5


∴ there are two cases


⇒ x + 2 > 5 and –(x + 2) < 5


⇒ x > 3 and –x – 2 < 5


⇒ x > 3 and –x > 7


⇒ x > 3 and x < -7


[Taking negative both sides invert the inequality sign]


(viii) ≤


-2x + 1 ≥ 9


Adding -1 both side, we get


⇒ -2x ≥ 8


⇒ x ≤ -4 [Dividing by negative inverts the inequality sign]


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