Number of Solutions - Linear Equations.

It is given that for some real numbers a and b, the system of equations $$x + y = 4$$ and $$(a+5)x+(b^2-15)y=8b$$ has infinitely many solutions for x and y.

Hence, we can say that 

=> $$\ \frac{\ a+5}{1}=\ \frac{\ b^2-15}{1}=\ \frac{\ 8b}{4}$$

This equation can be used to find the value of a, and b.

Firstly, we will determine the value of b. 

=> $$\ \frac{\ b^2-15}{1}=\ \frac{\ 8b}{4}\ =>\ b^2-2b-15=0$$

=> $$\left(b-5\right)\left(b+3\right)=0$$

Hence, the values of b are 5, and -3, respectively. 

The value of a can be expressed in terms of b, which is $$a+5=b^2-15\ =>\ a\ =b^2-20$$

When $$b=5,a=5^2-20=5$$

When $$b=-3,a=3^2-20=-11$$

The maximum value of $$ab=(-3)\cdot(-11)=33$$

The correct option is A

Arranging the equation, we know that there are no solutions when the lines are parallel:
for that the condition had to $$\frac{p}{3}=-\frac{4}{k}\ne\ \frac{2}{a}$$

Checking through options: 
Option A: using the first and last terms in our relation, we see that ap must not be equal to 6 for the lines to be parallel. This option puts no conditions on that and thus is not relevant. 

Option C: This question is the opposite of what we want; if this is true, the lines can never be parallel. 

Option D: Using the first and second terms of the relation, we see that we want kp = -12, or kp-12 = 0. Hence, this statement is not what we want. 

Option B: Using the second and third terms, we see that we do not want k equals to -2a, or we do not want k+2a to be equal to 0

Therefore, B is a condition that is necessary for the lines to be parallel and have no solution. 

Therefore, Option B is the correct answer. 

The maximum number of fruits one can pick without having 4 fruits of the same kind equals 9 when x=y=z=3.
So, if one picks 10 fruits, there will be at least 4 fruits of the same kind.