SOLUTION: Find the value of p for which the following system of eq has exactly 1 solution 7x-5y-4=0 and 14x+py+4=0

Question 709407: Find the value of p for which the following system of eq has exactly 1 solution 7x-5y-4=0 and 14x+py+4=0
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
p must be chosen just so that the slopes of the two lines are different. That condition is p<-10 and p>-10. p can be any real number, just not -10.

Your equations:
7x-5y-4=0 and 14x+py+4=0

The first one, -5y=-7x+4, y=(7/5)x+4, the slope is (7/5).
The second equation, py=-14x-4, y=(-14/p)x-4, the slope is variably (-14/p).
If the slopes of the two lines are DIFFERNT, then the system will have ONE solution. If p=-10, then the second line has slope (-14/-10)=7/5, and is parallel to the first line and then the system has no solution.

The only requirement for the two lines meeting in one point is that p NOT equal -10.
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