# Fill in the blank

(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 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]

-2x + 1 ≥ 9

Adding -1 both side, we get

-2x ≥ 8

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

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