Answer:
1. [tex](A,B) = (3,-2)[/tex]
2. The values of t are: -3, -1
Step-by-step explanation:
Given
[tex]\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}[/tex]
[tex]|t| = 2t + 3[/tex]
Required
Solve for the unknown
Solving [tex]\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}[/tex]
Take LCM
[tex]\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{(x - 2)(x-1)}[/tex]
Expand the denominator
[tex]\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{x^2 - 2x + x -2}[/tex]
[tex]\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{x^2 - x -2}[/tex]
Both denominators are equal; So, they can cancel out
[tex]x + 7 = A(x+1) + B(x-2)[/tex]
Expand the expression on the right hand side
[tex]x + 7 = Ax + A + Bx - 2B[/tex]
Collect and Group Like Terms
[tex]x + 7 = (Ax + Bx) + (A - 2B)[/tex]
[tex]x + 7 = (A + B)x + (A - 2B)[/tex]
By Direct comparison of the left hand side with the right hand side
[tex](A + B)x = x[/tex]
[tex]A - 2B = 7[/tex]
Divide both sides by x in [tex](A + B)x = x[/tex]
[tex]A + B = 1[/tex]
Make A the subject of formula
[tex]A = 1 - B[/tex]
Substitute 1 - B for A in [tex]A - 2B = 7[/tex]
[tex]1 - B - 2B = 7[/tex]
[tex]1 - 3B = 7[/tex]
Subtract 1 from both sides
[tex]1 - 1 - 3B = 7 - 1[/tex]
[tex]-3B = 6[/tex]
Divide both sides by -3
[tex]B = -2[/tex]
Substitute -2 for B in [tex]A = 1 - B[/tex]
[tex]A = 1 - (-2)[/tex]
[tex]A = 1 + 2[/tex]
[tex]A = 3[/tex]
Hence;
[tex](A,B) = (3,-2)[/tex]
Solving [tex]|t| = 2t + 3[/tex]
Because we're dealing with an absolute function; the possible expressions that can be derived from the above expression are;
[tex]t = 2t + 3[/tex] and [tex]-t = 2t + 3[/tex]
Solving [tex]t = 2t + 3[/tex]
Make t the subject of formula
[tex]t - 2t = 3[/tex]
[tex]-t = 3[/tex]
Multiply both sides by -1
[tex]t = -3[/tex]
Solving [tex]-t = 2t + 3[/tex]
Make t the subject of formula
[tex]-t - 2t = 3[/tex]
[tex]-3t = 3[/tex]
Divide both sides by -3
[tex]t = -1[/tex]
Hence, the values of t are: -3, -1
