Question Video: Solving Rational Equations

Given that 𝑛₁(π‘₯) = (7 + 𝑏)/(π‘₯ βˆ’ 7), 𝑛₂(π‘₯) = 2/(π‘₯ βˆ’ 7), and 𝑛₁(π‘₯) = 𝑛₂(π‘₯), what is the value of 𝑏?

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Video Transcript

Given that 𝑛 sub one of π‘₯ is equal to seven plus 𝑏 over π‘₯ minus seven, 𝑛 sub two of π‘₯ is equal to two over π‘₯ minus seven, and 𝑛 sub one of π‘₯ is equal to 𝑛 sub two of π‘₯, what is the value of 𝑏?

Here, we have two rational functions. And we’re told that the two functions are equal to each other. So that means that for 𝑛 sub one of π‘₯ to be called 𝑛 sub two of π‘₯, the functions are equal for all values of π‘₯, not just for a certain value or values. So we can say that seven plus 𝑏 over π‘₯ minus seven equals two over π‘₯ minus seven. Now, look really carefully. The denominator of each of our functions is equal. And so for the function 𝑛 sub one of π‘₯ to be equal to the function 𝑛 sub two of π‘₯, their numerator must also be equal. So seven plus 𝑏 must be equal to two.

Now, we have this; we can solve this equation for 𝑏 fairly easily. We’re going to subtract seven from both sides. And seven plus 𝑏 minus seven is 𝑏. And two minus seven is negative five. And so we found the value of 𝑏; it’s negative five. Now, of course, we can check that what we’ve done is correct by substituting 𝑏 into our expression for the function 𝑛 sub one of π‘₯. We get seven plus negative five over π‘₯ minus seven, which is two over π‘₯ minus seven as required.

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