Question

State any excluded values.

2(a-1)/3(a+1)

Answer 1

When a denominator is 0...we have an undefined function because we cannot divide by 0

So all you need to do is find out when

\[\large 3(a + 1) = 0\]

what value of 'a' makes that equation = 0?

Answer 2

idk negative something?

Answer 3

Right...

3(a + 1) = 0

Well 3 is being multiplied by those parenthesis...any we know we want 0...well anything times 0 = 0 ...so lets make those parenthesis = 0

(a + 1) = 0

Must mean a = -1 right?

Answer 4

3(a+1)=0 ;so the ans:a=-1

Answer 5

So on my paper do i just write: 3(a+1)=0 ;so the ans:a=-1