Question Video: Finding the Domain for Two Rational Functions to Be Equal | Nagwa Question Video: Finding the Domain for Two Rational Functions to Be Equal | Nagwa

# Question Video: Finding the Domain for Two Rational Functions to Be Equal Mathematics • Third Year of Preparatory School

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Given the functions πβ(π₯) = π₯/(π₯Β² β 10π₯) and πβ(π₯) = 1/(π₯ β 10), what is the set of values on which πβ = πβ?

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### Video Transcript

Given the functions π one of π₯ is equal to π₯ over π₯ squared minus 10π₯ and π two of π₯ is equal to one over π₯ minus 10, what is the set of values on which π one is equal to π two?

Weβre given two rational functions π one and π two, where a rational function is a function of the form π of π₯ over π of π₯, such that π and π are polynomials in π₯. And the domain of the function, thatβs the input values π₯, are such that π of π₯ is not equal to zero. Now, weβre asked to find the set of values on which π one is equal to π two, that is, the domain or set of input values satisfying π one equals π two. This means we need to exclude any values of π₯ that make our denominators equal to zero. And so we need to solve π₯ squared minus 10π₯ is equal to zero and π₯ minus 10 is equal to zero.

Letβs look first at π one, where we have a common factor of π₯ in both terms. And taking this outside some parentheses, we have π₯ multiplied by π₯ minus 10 is equal to zero. Solutions to this are either π₯ is equal to zero or π₯ minus 10 is equal to zero. And to solve π₯ minus 10 is equal to zero, we add 10 to both sides, giving us π₯ is equal to 10. Our solutions are therefore π₯ is equal to zero and π₯ is equal to 10. The domain of π one of π₯ is therefore the set of real numbers not including the set containing zero and 10.

And now turning to π two of π₯, we see that the denominator is equal to zero if π₯ is equal to 10. And so the domain of π two of π₯ is the set of real numbers not including the set containing the number 10.

Now letβs take a closer look at the function π one of π₯. Weβve seen that we have a common factor of π₯ in the denominator, which we can take outside some parentheses. If we then divide both numerator and denominator by π₯, π one is equal to one over π₯ minus 10, which is actually equal to our second function π two of π₯. This means that the set of values on which π one is equal to π two is the domain of π one. And this is because the domain of π one includes the domain of π two and our domain must apply to both functions. The set of values on which π one is equal to π two is therefore the set of real numbers not including the set containing the numbers zero and 10.

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