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Hi M, as you know the domain is a set of x values involved in a graph. x -3 and 5x-3, alone, are linear and any number you put in for x, you will get a y which means the graph exist at that point. Any equation that is linear(y=mx + b) or quadratic (y=x2), cubic,...,with no denominator is the set of all x values because any value you put in the equation will have a y-value. However when (x-3)/(5x-3), there is a value of x where it would make the denominator 0, which would make the graph undefined at that point. This is why, in finding the domain, you set 5x-3 not= 0. There will be a vertical asymptote at that point that the graph will get close to, but never touch. If you have a radical such as the sqrt(x-4) or (x-4)^1/2, square roots cannot be less than 0 or negative, if they are, they are not real numbers. To solve sqrt(x-4), simply set x-4 greater than or equal to 0. Another scenario to look out for is square roots in denominator-you might be tempted to set the square root to greater than or equal to 0, but you cannot have a 0 in the denominator so you would set the square root to greater than 0.
In summary, denominators can NEVER equal zero and square roots, with real numbers, cannot be less than zero.
Best of luck,