Given that x = 4, y = 2 is a solution of the simultaneous equations ax + by = 8 and b^2x + a^2y= 34, find the possible values of a and b?

Answer 1

To find the possible values of a and b, we can substitute the given values of x = 4, y = 2 into the two equations:

1. ax + by = 8
a(4) + b(2) = 8
4a + 2b = 8
2a + b = 4

2. b^2x + a^2y= 34
b^2(4) + a^2(2) = 34
4b^2 + 2a^2 = 34
2b^2 + a^2 = 17
b^2 = 17 - a^2

Now substitute 2a + b = 4 into b^2 = 17 - a^2:

b^2 = 17 - a^2
(2a + b)^2 = 17 - a^2
(2a)^2 + 2(2a)b + b^2 = 17 - a^2
4a^2 + 4ab + b^2 = 17 - a^2
5a^2 + 4ab + b^2 - 17 = 0
a^2 + 4a + b - 3 = 0

Using x = 4, y = 2 as a solution:

a^2 + 4a + b - 3 = 0
4^2 + 4(4) + 2 - 3 = 0
16 + 16 + b - 3 = 0
32 + b - 3 = 0
b + 29 = 0
b = -29

Now substitute b = -29 into 2a + b = 4:

2a + (-29) = 4
2a - 29 = 4
2a = 33
a = 33/2
a = 16.5

Therefore, the only solution for a and b that satisfies the given simultaneous equations is a = 16.5 and b = -29.