The anwser for this question is B
Answer:
A and C
Step-by-step explanation:
Wait this is tricky let me go ask my sis she is coming the answer is in the comeent!
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

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)
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Given

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 = 
x =
and x = - 2 are excluded values → (a)
4x + 2 = 62
4x = 62 - 2
4x = 60
x = 15
12y = 144
y = 12
A is the answer.