Question 3.

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

=> (b - 5)(b + 3) = 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)⋅(−11) = 33