Answer:
(a) and (a)
Step-by-step explanation:
In both questions the denominator of the rational functions cannot be zero as this would make them undefined. Equating the denominators to zero and solving gives the values that x cannot be.
Given
[tex]\frac{x-3}{(3-x)(2+x)}[/tex]
solve (3 - x)(2 + x) = 0
Equate each factor to zero and solve for x
3 - x = 0 ⇒ x = 3
2 + x = 0 ⇒ x = - 2
x = 3 and x = - 2 are excluded values → (a)
------------------------------------------------------------------------
Given
[tex]\frac{-9x+3}{6x^2+10x-4}[/tex]
solve
6x² + 10x - 4 = 0 ← in standard form
(x+ 2)(6x - 2) = 0 ← in factored form
Equate each factor to zero and solve for x
x + 2 = 0 ⇒ x = - 2
6x - 2 = 0 ⇒ 6x = 2 ⇒ x = [tex]\frac{1}{3}[/tex]
x = [tex]\frac{1}{3}[/tex] and x = - 2 are excluded values → (a)
