How do i solve this? can you show me step by step process.

(Q14) To find the equation when given the gradient and one set of points (or coordinates), you have to solve for b. 
1. Identify the gradient and x- and y- intercepts. 
    m=3, x=1, y=5
2. Substitute them into the equation, y=mx+b. 
    5=3(1) + b 
3. Solve for b.
    5=3 + b 
    b=2 
4. Rewrite the equation. 
   y=3x+ 2
(Q15) To find the equation when given two sets of points, you need to.
1. Find the gradient, using the formula, (x2-x1 / y2-y1) 
   (7,5) and (4, -7)
    x1=7, x2= 4, y1=5, y2=-7
    4-7/-7-5 = -3/-12 = 1/4 
   Therefore, gradient = 1/4 
2. Substitute with one set of points (can be either (7,5) or (4,-7) in this example - question 15) into the equation, (like you did in question 14) 
  a) Identify the gradient and x- and y- intercepts. 
   m=1/4, x=7, y=5
  b) Solve for b. 
   5=1/4(7) + b
   5=7/4 + b 
   b=13/4 or 3 1/4 
  c) Rewrite the equation.
   y=1/4x + 13/4

The times are t = 9, t = 10, t = 11 and t = 12

Step-by-step explanation:

For the train to be more than 16 km South and since south is taken as negative,

d(t) > -16

t³ - 9t² + 6t > -16

t³ - 9t² + 6t + 16 > 0

Since -1 is a factor of 16, inserting t = -1 into the d(t), we have

d(-1) = (-1)³ - 9(-1)² + 6(-1)+ 16 = -1 - 9 - 6 + 16 = -16 + 16 = 0. By the factor theorem, t + 1 is a factor of d(t)

So, d(t)/(t + 1) = (t³ - 9t² + 6t + 16)/(t +1) = t² - 10t + 16

Factorizing  t² - 10t + 16, we have

t² - 2t - 8t + 16

= t(t - 2) - 8(t - 2)

= (t -2)(t - 8)

So t - 2 and t - 8 are factors of d(t)

So (t + 1)(t -2)(t - 8) > 0

when t < -1, example t = -2 ,(t + 1)(t -2)(t - 8) = (-2 + 1)(-2 -2)(-2 - 8) = (-1)(-4)(-10) = -40 < 0

when -1 < t < 2, example t = 0 ,(t + 1)(t -2)(t - 8) = (0 + 1)(0 -2)(0 - 8) = (1)(-2)(-8) = 16 > 0

when 2 < t < 8, example t = 3 ,(t + 1)(t -2)(t - 8) = (3 + 1)(3 -2)(3 - 8) = (4)(1)(-5) = -20 < 0

when t > 8, example t = 9,(t + 1)(t -2)(t - 8) = (9 + 1)(9 -2)(9 - 8) = (10)(7)(1) = 70 > 0

Since t cannot be negative, d(t) is positive in the interval 0 < t < 2 and t > 8

Since t ∈ (0, 12]

In the interval 0 < t < 2 the only value possible for t is t = 1

d(1) = t³ - 9t² + 6t = (1)³ - 9(1)² + 6(1) = 1 - 9 + 6 = -2

Since d(1) < -16 this is invalid

In the interval t > 8 , the only possible values of t are t = 9, t = 10.t = 11 and t = 12.

So,

d(9) = 9³ - 9(9)² + 6(9) = 0 + 54 = 54 km

d(10) = 10³ - 9(10)² + 6(10) = 1000 - 900 + 60 = 100 +60 = 160 km

d(11) = 11³ - 9(11)² + 6(11) = 1331 - 1089 + 66 = 242 + 66 = 308 km

d(12) = 12³ - 9(12)² + 6(12) = 1728 - 1296 + 72 = 432 + 72 = 504 km