Question

6E and 7 with full explanation

Answer 1

See below ~

Explanation:

Question 6(e) :

⇒ x² ≤ 16

Take square root on each side :

⇒ √x² ≤ √16

⇒ x ≤ 4 and x ≥ -4

⇒ -4 ≤ x ≤ 4

Answer 2

Part 6e

⇒ x² ≤ 16

Apply the absolute rule if x² < a then -√a < x < √a

⇒ -√16 ≤ x ≤ +√16

⇒ -4 ≤ x ≤ 4

Part 7

First of all, The student should have subtracted both sides by 4

What he should have done:

`\rightarrow \sf (3)/(x-2) > 4`

`\rightarrow \sf (3)/(x-2) -4 > 4 -4`

`\rightarrow \sf (3)/(x-2) -(4(x-2))/(x-2) > 0`

`\rightarrow \sf (3-4(x-2))/(x-2)} > 0`

`\rightarrow \sf (3-4x+8)/(x-2)} > 0`

`\rightarrow \sf (-4x+11)/(x-2)} > 0`

`\sf \large \boxed{\sf {\left \{ {{-4x+11 = 0} \atop {x-2 = 0}} \right. }}`

⇒ -4x + 11 = 0, x - 2 = 0

⇒ -4x = -11, x = 2

⇒ x = 11/4 , x = 2

Solution satisfying the inequality:

⇒ 2 < x < 11/4