How do you subtract #(x^2+9)/(x-3)- (6x)/(x-3)#?

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Answer 1 · 2017-07-16T14:07:31

See a solution process below:

Because both fractions have the same denominator we can subtract the numerators over the common denominator:

#(x^2 + 9)/(x - 3) - (6x)/(x - 3) => (x^2 + 9 - 6x)/(x - 3) =>#

#(x^2 - 6x + 9)/(x - 3)#

We can now factor the numerator as:

#((x - 3)(x - 3))/(x - 3)#

Now, we can cancel common terms in the numerator and denominator:

#(color(red)(cancel(color(black)((x - 3))))(x - 3))/color(red)(cancel(color(black)(x - 3))) =>#

#x - 3#

However, we must remember we cannot divide by zero. Therefore:

#x - 3 != 0# Or #x != 3#