Question

1. If (a,b) is a solution of the system of equations {P=0 and Q=0}, show that it must also be a solution of the system of equations {P=0 and Q+kP=0}.
2. If (a,b) is a solution of the system of equations {P=0 and Q+kP=0}, show that it must also be a solution of the system of equations {P=0 and Q=0}.

Answer 1

Where are you stuck?

Answer 2

1) if (a,b) is a solution of (P=0 and Q =0) , that is
P(a,b) =0 and Q (a,b) =0
since P(a,b) =0, k*P(a,b) =0 =k*0=0

Answer 3

I don't even know where to begin. I am completely lost.

Answer 4

on the other hand (Q + kP =0) means (Q+kP) (a,b) =0 . this is Q+ kP "apply on (a,b), it is NOT (Q+kP)* (a,b)

Answer 5

You must understand this concept to step up

Answer 6

Thank you, but I just wanted to know what would the final answer be?

Answer 7

For example: P(x) = x^2-1 , then if x =1 is a solution of P(x), then P(1) =0

Answer 8

YOur problem is a "Show" problem, it doesn't have final answer.

Answer 9

You have to do some logic to get the credit. In other words, you have to PROVE it

Answer 10

I give you the whole proof for the first one, you do the second one, OK?

Answer 11

Again, thank you.

Answer 12

(a,b) is a solution of P=0 and Q =0 (given) . That is P(a,b) =0 , hence k*P(a,b) =k*P(a,b) =0
hence \(Q(a,b) =0 ~~(given)\\k*P(a,b) =0~~(above)\\-------------------\\Q(a,b)+kP(a,b) =0+0=0\\(Q+kP)(a,b)=0\)
that shows (a,b) is a solution of \((Q+kP)\)
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