Answer:
1. ![(A,B) = (3,-2)](https://tex.z-dn.net/?f=%28A%2CB%29%20%3D%20%283%2C-2%29)
2. The values of t are: -3, -1
Step-by-step explanation:
Given
![\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%20%2B%207%7D%7Bx%5E2%20-%20x%20-%202%7D%20%3D%20%5Cfrac%7BA%7D%7Bx%20-%202%7D%20%2B%20%5Cfrac%7BB%7D%7Bx%20%2B%201%7D)
![|t| = 2t + 3](https://tex.z-dn.net/?f=%7Ct%7C%20%3D%202t%20%2B%203)
Required
Solve for the unknown
Solving ![\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%20%2B%207%7D%7Bx%5E2%20-%20x%20-%202%7D%20%3D%20%5Cfrac%7BA%7D%7Bx%20-%202%7D%20%2B%20%5Cfrac%7BB%7D%7Bx%20%2B%201%7D)
Take LCM
![\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{(x - 2)(x-1)}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%20%2B%207%7D%7Bx%5E2%20-%20x%20-%202%7D%20%3D%20%5Cfrac%7BA%28x%2B1%29%20%2B%20B%28x-2%29%7D%7B%28x%20-%202%29%28x-1%29%7D)
Expand the denominator
![\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{x^2 - 2x + x -2}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%20%2B%207%7D%7Bx%5E2%20-%20x%20-%202%7D%20%3D%20%5Cfrac%7BA%28x%2B1%29%20%2B%20B%28x-2%29%7D%7Bx%5E2%20-%202x%20%2B%20x%20-2%7D)
![\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{x^2 - x -2}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%20%2B%207%7D%7Bx%5E2%20-%20x%20-%202%7D%20%3D%20%5Cfrac%7BA%28x%2B1%29%20%2B%20B%28x-2%29%7D%7Bx%5E2%20-%20x%20-2%7D)
Both denominators are equal; So, they can cancel out
![x + 7 = A(x+1) + B(x-2)](https://tex.z-dn.net/?f=x%20%2B%207%20%3D%20A%28x%2B1%29%20%2B%20B%28x-2%29)
Expand the expression on the right hand side
![x + 7 = Ax + A + Bx - 2B](https://tex.z-dn.net/?f=x%20%2B%207%20%3D%20Ax%20%2B%20A%20%2B%20Bx%20-%202B)
Collect and Group Like Terms
![x + 7 = (Ax + Bx) + (A - 2B)](https://tex.z-dn.net/?f=x%20%2B%207%20%3D%20%28Ax%20%2B%20Bx%29%20%20%2B%20%28A%20-%202B%29)
![x + 7 = (A + B)x + (A - 2B)](https://tex.z-dn.net/?f=x%20%2B%207%20%3D%20%28A%20%2B%20B%29x%20%2B%20%28A%20-%202B%29)
By Direct comparison of the left hand side with the right hand side
![(A + B)x = x](https://tex.z-dn.net/?f=%28A%20%2B%20B%29x%20%3D%20x)
![A - 2B = 7](https://tex.z-dn.net/?f=A%20-%202B%20%3D%207)
Divide both sides by x in ![(A + B)x = x](https://tex.z-dn.net/?f=%28A%20%2B%20B%29x%20%3D%20x)
![A + B = 1](https://tex.z-dn.net/?f=A%20%2B%20B%20%3D%201)
Make A the subject of formula
![A = 1 - B](https://tex.z-dn.net/?f=A%20%3D%201%20-%20B)
Substitute 1 - B for A in ![A - 2B = 7](https://tex.z-dn.net/?f=A%20-%202B%20%3D%207)
![1 - B - 2B = 7](https://tex.z-dn.net/?f=1%20-%20B%20-%202B%20%3D%207)
![1 - 3B = 7](https://tex.z-dn.net/?f=1%20-%203B%20%3D%207)
Subtract 1 from both sides
![1 - 1 - 3B = 7 - 1](https://tex.z-dn.net/?f=1%20-%201%20-%203B%20%3D%207%20-%201)
![-3B = 6](https://tex.z-dn.net/?f=-3B%20%3D%206)
Divide both sides by -3
![B = -2](https://tex.z-dn.net/?f=B%20%3D%20-2)
Substitute -2 for B in ![A = 1 - B](https://tex.z-dn.net/?f=A%20%3D%201%20-%20B)
![A = 1 - (-2)](https://tex.z-dn.net/?f=A%20%3D%201%20-%20%28-2%29)
![A = 1 + 2](https://tex.z-dn.net/?f=A%20%3D%201%20%2B%202)
![A = 3](https://tex.z-dn.net/?f=A%20%3D%203)
Hence;
![(A,B) = (3,-2)](https://tex.z-dn.net/?f=%28A%2CB%29%20%3D%20%283%2C-2%29)
Solving ![|t| = 2t + 3](https://tex.z-dn.net/?f=%7Ct%7C%20%3D%202t%20%2B%203)
Because we're dealing with an absolute function; the possible expressions that can be derived from the above expression are;
and ![-t = 2t + 3](https://tex.z-dn.net/?f=-t%20%3D%202t%20%2B%203)
Solving ![t = 2t + 3](https://tex.z-dn.net/?f=t%20%3D%202t%20%2B%203)
Make t the subject of formula
![t - 2t = 3](https://tex.z-dn.net/?f=t%20-%202t%20%3D%203)
![-t = 3](https://tex.z-dn.net/?f=-t%20%3D%203)
Multiply both sides by -1
![t = -3](https://tex.z-dn.net/?f=t%20%3D%20-3)
Solving ![-t = 2t + 3](https://tex.z-dn.net/?f=-t%20%3D%202t%20%2B%203)
Make t the subject of formula
![-t - 2t = 3](https://tex.z-dn.net/?f=-t%20-%202t%20%3D%203)
![-3t = 3](https://tex.z-dn.net/?f=-3t%20%3D%203)
Divide both sides by -3
![t = -1](https://tex.z-dn.net/?f=t%20%3D%20-1)
<em>Hence, the values of t are: -3, -1</em>