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
