Write a linear function that passes through (1,5) and (2,15)
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namely, let's rationalize the denominator in the fraction, for which case we'll be using the <u>conjugate</u> of that denominator, so we'll multiply top and bottom by its <u>conjugate</u>.
so the denominator is 5 + i, simply enough, its conjugate is just 5 - i, recall that same/same = 1, thus (5-i)/(5-i) = 1, and any expression multiplied by 1 is just itself, so we're not really changing the fraction per se.