Answer:
Remember that the division by zero is not defined, this is the criteria that we will use in this case.
1) [tex]\frac{y^2 - 1}{y} + \frac{y}{y - 3}[/tex]
So the fractions are defined such that the denominator is never zero.
For the first fraction, the denominator is zero when y = 0
and for the second fraction, the denominator is zero when y = 3
Then the fractions exist for all real values except for y = 0 or y = 3
we can write this as:
R / {0} U { 3}
(the set of all real numbers except the elements 0 and 3)
2) [tex]\frac{b + 4}{b^2 + 7}[/tex]
Let's see the values of b such that the denominator is zero:
b^2 + 7 = 0
b^2 = -7
b = √-7
This is a complex value, assuming that b can only be a real number, there is no value of b such that the denominator is zero, then the fraction is defined for every real number.
The allowed values are R, the set of all real numbers.
3) [tex]\frac{a}{a*(a - 1) - 1}[/tex]
Again, we need to find the value of a such that the denominator is zero.
a*(a - 1) - 1 = a^2 - a - 1
So we need to solve:
a^2 - a - 1 = 0
We can use the Bhaskara's formula, the two values of a are given by:
[tex]a = \frac{-(-1) \pm \sqrt{(-1)^2 + 4*1*(-1)} }{2*1} = \frac{1 \pm \sqrt{5} }{2}[/tex]
Then the two values of a that are not allowed are:
a = (1 + √5)/2
a = (1 - √5)/2
Then the allowed values of a are:
R / {(1 + √5)/2} U {(1 - √5)/2}
