Answer :

(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]

Rate this question :

Solve each of theRD Sharma - Mathematics

A solution is to RD Sharma - Mathematics

A company manufacRD Sharma - Mathematics

<span lang="EN-USRS Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal - Mathematics

Solve each of theRS Aggarwal - Mathematics

Solve x + 5 > RS Aggarwal - Mathematics

Find the solutionRS Aggarwal - Mathematics

If |x – 1| > 5Mathematics - Exemplar