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.