SOLUTION: Find the values of A and B if, A/z+2 + B/2z-3= 5z-11/2z^2+z-6

Question 110404This question is from textbook Glencoe Algebra 2 Integration Applications Connections
: Find the values of A and B if,
A/z+2 + B/2z-3= 5z-11/2z^2+z-6 This question is from textbook Glencoe Algebra 2 Integration Applications Connections

Answer by TP(29) (Show Source): You can put this solution on YOUR website!
A/z+2 + B/2z-3 = [A(2z-3) + B(z+2)]/(z+2)(2z-3)
= (2Az-3A + Bz+2B)/(2z^2-3z+4z-6)
= (2Az+Bz-3A+2B)/(2z^2+z-6)
But we are given that this is equal to (5z-11)/(2z^2+z-6) and since the denominators(the bottom part of each expression) are equal then it follows that the numerators(the top part of each expression) are also equal.
Hence
2Az+Bz-3A+2B = 5z-11.
The left hand side can be written as (2A+B)z-(3A-2B) [remember - sign in front of a bracket means that the sign inside changes so that -3A+2B= -(3A-2B)].
And so we now have:
(2A+B)z-(3A-2B) = 5z - 11.
This must mean that 2A+B = 5 (i) and 3A-2B = 11 (ii) [This process is called equating the coefficients.]
Now multiply (i) by 2 [in order to get 2B] and (i) becomes:
4A+2B = 10 (i)
Now by adding this new form of equation (i) to equation (ii) we get:
7A +0 = 21 [4A+3A,2B add -2B and 10+11].
So 7A = 21
Divide both sides by 7:
A = 3 ANS
To find B replace A by 3 in the original form of equation (i) and we get:
2*3 + B = 5
Hence 6 + B = 5
Subtract 6 from both sides and we get:
B = -1 ANS
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