Question 47

# 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. Then, the maximum possible value of ab is

Solution

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

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