Algebra
posted by y912f .
91. Find the domain of: y=3x+1
a. all negative numbers
b. x>0
c. all real numbers
d. x=3
Can someone please show the work and how to solve this problem. There are many of these in my lessons and I don't understand how to solve them. Do you solve for x?

The domain is all values of x for which the equation "makes sense" and can be used to calculate y. In this case, the answer is c (all real numbers).

No. You just inspect the function (in this case (3x+1) and see if there are any values of x that would violate the fundamental laws of math. For instance, if you had (3x+1)/(x+3) then the domain would be all values of x EXCEPT x=3, as division by zero is NOT allowed. In your case, the domain of x is (best answer: all real numbers). X can assume any value (real or imaginary), but I suppose you are not into complex numbers yet. Let me let you jump ahead and look at that(I am so kind and generous).
What if f(x)=x/(x^2+1) Well, division by zero is not allowed, so x= sqrt(1) is not allowed, but that is an imaginary number. So the domain for x here is all values, real and imaginary, except x=i (i=sqrt(1))
In your problem above, all real x are allowed.
Here is a good explanation: http://www.analyzemath.com/DomainRange/DomainRange.html