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#### Fill in the blanks of the following: (i) If – 4x ≥ 12, then x ... – 3. (ii) If -3x/4 ≤ –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

-x ≥ 3     ……….. (Dividing by 4)

Now, we will invert the equation by taking negative signs on both the sides, we get,

x ≤ 3

(ii)  -3/4 x ≤ -3

Thus, -3x ≤ -12     …… (On multiplying by 4)

Now, we will invert the equation by taking negative signs on both the sides, we get,

3x ≤ 12 I.e., x ≤ 4.

(iii) 2/x+2 > 0

In order for the above to be greater than 0,

x+2 > 0 i.e., x > -2

(iv) x > -5

I.e., 4x > -20    …….. (on multiplying by 4)

(v) x > y

z is negative, since z < 0

Now, we will invert the equation by taking negative signs on both the sides, we get,

Thus, -xz > - yz

(vi) q < 0

Now, we will invert the equation by taking negative signs on both the sides, we get,

-q > 0

p - q > p ….. (adding p on both sides)

(vii) |x+2|> 5

Thus, there will be two cases,

X+2 > 5 & -(x+2)<5

X>3 & -x-2 <5

X > 3 & -x > 7

X > 3 & x < -7

(viii) -2x + 1 ≥ 0

-2x ≥ 8  …… (on adding -1 to both sides)

Now, we will invert the equation by taking negative signs on both the sides, we get,

Thus, x ≤ -4.