Question 905083
First solve for y


{{{xy + 3y = 2x + 1}}}


{{{y(x + 3) = 2x + 1}}}


{{{y = (2x + 1)/(x + 3)}}}


The function is {{{y = (2x + 1)/(x + 3)}}} which can also be written as {{{f(x) = (2x + 1)/(x + 3)}}} (since y = f(x))


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You cannot divide by zero. So {{{x+3}}}, the denominator, can't be equal to zero. Set it equal to zero and solve for x


{{{x+3=0}}}


{{{x=-3}}}


The value {{{x=-3}}} causes the denominator to be zero. Therefore, we must kick it out of the domain. Any other real number is perfectly fine since it doesn't cause a division by zero error.


Domain: Set of all real numbers x such that {{{x<>-3}}}


Domain in interval notation: *[Tex \Large (-\infty,-3)\cup(-3,\infty)]