Question

For what values of the variable, do the following fractions exist: y^2-1/y+y/y-3

Answer 1

Answer:

Remember that the division by zero is not defined, this is the criteria that we will use in this case.

1) $\frac{y^2 - 1}{y} + \frac{y}{y - 3}$

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) $\frac{b + 4}{b^2 + 7}$

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) $\frac{a}{a*(a - 1) - 1}$

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:

$a = \frac{-(-1) \pm \sqrt{(-1)^2 + 4*1*(-1)} }{2*1} = \frac{1 \pm \sqrt{5} }{2}$

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}