For what values of the variable, do the following fractions exist: y^2-1/y+y/y-3 For what values of the variable, do the following fractions exist: b+4/b^2+7

For what values of the variable, do the following fractions exist: a/a(a-1)-1

PLEASE HELP NEED ANSWER ASA

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

1)

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)

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)

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:

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}

step-by-step explanation:

my butt

a=24; b=12

step-by-step explanation:

a/b=2: 1

a=2b

36/(1+2)=12

b=12

a=24