Pre Calculus

Answer for Q1: -1.14

Solution:

sqrt(x + 3y) + sqrt(2xy) = 9.72
d[sqrt(x + 3y) + sqrt(2xy)] / dx = d(9.72) / dx
[1 + 3(dy/dx)] / [2sqrt(x + 3y)] + [y + x(dy/dx)] / sqrt(2xy) = 0
[1 + 3(dy/dx)][sqrt(2xy)] + [y + x(dy/dx)][2sqrt(x + 3y)] = 0

sub. x = 3, y = 5, we have
[1 + 3(dy/dx)]{sqrt[2(3)(5)]} + [5 + 3(dy/dx)][2sqrt(3 + 3(5))] = 0
[1 + 3(dy/dx)](sqrt30) + [5 + 3(dy/dx)](2sqrt18) = 0
sqrt30 + 3(sqrt30)(dy/dx) + 5(2sqrt18) + 3(2sqrt18)(dy/dx) = 0
sqrt30 + 3(sqrt30)(dy/dx) + 30sqrt2 + 18sqrt2(dy/dx) = 0
3(sqrt30)(dy/dx) + 18sqrt2(dy/dx) = - (sqrt30) - 30sqrt2
[3(sqrt30) + 18sqrt2](dy/dx) = - (sqrt30) - 30sqrt2
dy/dx = [- (sqrt30) - 30sqrt2] / [3(sqrt30) + 18sqrt2] = -1.14



Answer for Q2: a = -1, b = -3

Solution:

[-34, 7, -27] = a[1, -1, 3] + b[11, -2, 8]
[-34, 7, -27] = [a, -a, 3a] + [11b, -2b, 8b]
[-34, 7, -27] = [a + 11b, -a - 2b, 3a + 8b]
thus, we have
a + 11b = -34 ------- (i)
-a - 2b = 7 ----------- (ii)
3a + 8b = -27 ------- (iii)

(i) + (ii) => 9b = -27 => b = -3
sub b = -3 into (i) => a + 11(-3) = -34 => a = -1
sub a = -1 and b = -3 into (iii) for checking, LHS = 3(-1) + 8(-3) = -27 = RHS
therefore, a = -1, b = -3