how does k=0 still make and expression undefined when there are still whole numbers

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Helen Window

16 de fevereiro de 2016 08:43

− 10k+13/6k+4 This fraction is the simplified version of an expression after the monomial factors have been cancelled out. The values of k that make the expression undefined are k= 0 and k=-2/3. The note at the end of the answer says Note that k= 0 still makes the expression undefined, even in its simplified form.
I dont understand this answer : if 6k = 0 then 0+4 is still 4 and -0+13 is still 13, can you help me understand this please.

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KooRee

28 de fevereiro de 2016 05:49
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Do you mean (-10k+13)/(6k+4)? or is one term -10k, another term (13/6)*k or 13/(6k), and the last term 4? I'm going to assume you mean (-10k+13)/(6k+4) based on what you said about monomial factors cancelling out.

When you simplify an expression, you are finding an EQUIVALENT expression, but it's not EXACTLY the same. You still have to take into account the original expression.

So, for example, if the original problem was [(k)(-10k+13)] / [(k)(6k+4)], then think about what would happen if k=0?
Just looking at the denominator, (k)(6k+4), it would become (0)(6*0+4), and, of course, anything times 0 is 0, so it would make our WHOLE denominator 0, which is undefined.

So even though you can cancel the individual factors, you still have to take into account the original expression. Hope this helps.
Helen Window

29 de fevereiro de 2016 11:56
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so obvious when you put it like that. That was very helpful, thank you.

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