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6E and 7 with full explanation

6E And 7 With Full Explanation class=

Sagot :

Answer:

See below ~

Step-by-step explanation:

Question 6(e) :

⇒ x² ≤ 16

Take square root on each side :

⇒ √x² ≤ √16

⇒ x ≤ 4 and x ≥ -4

⇒ -4 ≤ x ≤ 4

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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:

[tex]\rightarrow \sf \dfrac{3}{x-2} > 4[/tex]

[tex]\rightarrow \sf \dfrac{3}{x-2} -4 > 4 -4[/tex]

[tex]\rightarrow \sf \dfrac{3}{x-2} -\dfrac{4(x-2)}{x-2} > 0[/tex]

[tex]\rightarrow \sf \dfrac{3-4(x-2)}{x-2}} > 0[/tex]

[tex]\rightarrow \sf \dfrac{3-4x+8}{x-2}} > 0[/tex]

[tex]\rightarrow \sf \dfrac{-4x+11}{x-2}} > 0[/tex]

[tex]\sf \large \boxed{\sf {\left \{ {{-4x+11 = 0} \atop {x-2 = 0}} \right. }}[/tex]

⇒ -4x + 11 = 0, x - 2 = 0

⇒ -4x = -11, x = 2

⇒ x = 11/4 , x = 2

Solution satisfying the inequality:

⇒ 2 < x < 11/4

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