Kay solves the following system of equations using the substitution method.
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47 -14i
You can work this out in the straight-forward way, or you can recognize that (6-i) is a common factor. In the latter case, you have ...
... = (6-i)(5 + 3-i)
... = (6 -i)(8 -i)
This product of binomials is found in the usual way. Each term of one factor is multiplied by each term of the other factor and the results summed. Of course, i = √-1, so i² = -1.
... = 6·8 -6i -8i +i²
... = 48 -14i -1
... =
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A suitable graphing calculator will work these complex number problems easily.
