Question

given that z=(16-q)/(7-p)
find the possible values of p and q that has:
a) no solutions
b) infinite solutions
c) a unique solution

Answer 1

a) p,q,z are elements of N
b) p,q,z are elements of R
c) p,q,z are alements of Q

(not sure if that is right)

Answer 2

z=(16-q)/(7-p)
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a) If p = 7 and q is not 16, the denominator of the rational expression becomes 0, and there is no solution.

b) If p = 7 and q = 16, the fraction becomes 0/0 which is indeterminate.

c) If p is not 7, and if p and q are Real Numbers, I think there will be a unique solution.

I am not certain that these thoughts are correct. Someone else will come along and add other ideas, I hope.

@UnkleRhaukus

Answer 3

@Directrix you are right~

Answer 4

@Kystal --> Thanks for letting me know.

Answer 5

Do you know why 0/0 is indeterminate? I thought 0/0 gives 0?

Answer 6

0/0 does equal 0 but it could equal any other real number, too. Think about how division is defined in terms of multiplication. 6 divided by 2 = 3 if 3*2 =6. What happens with 0/0, I'll draw.

Does that make sense?