Which of the solution sets is all real numbers?
|x| < -1
|x| = -1
|x| > -1

The solution sets is all real numbers in case of:

                        |x| > -1

We know that modulus is a function with the property such that:

                      if a<0 then |a|= -a

that is the modulus of a negative number is positive and if a≥0

                                then |a| =a

and modulus of a positive value is also positive.

i.e. modulus function always gives positive value.

Hence,

1)

|x|<-1

This is not possible as modulus function always gives a value ≥0 for all real numbers.

2)

|x|= -1

This is also not possible as modulus of any number can't be negative.

3)

|x| > -1

The modulus of any number will definitely be greater than or equal to zero.

    Hence, the solution set contain all the real numbers.

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?