Question

Please help ASAP. The question is down below.

Answer 1

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

$\frac{x-3}{(3-x)(2+x)}$

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

$\frac{-9x+3}{6x^2+10x-4}$

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 = $\frac{1}{3}$

x = $\frac{1}{3}$ and x = - 2 are excluded values → (a)