Chapter 8 Systems of Equations and Inequalities

Chapter 8: Systems of Equations and Inequalities
33. Find the partial fraction decomposition:
x x A B
= = +
2
x + 2x−3 ( x + 3)( x− 1) x+ 3 x−1
Multiplying both sides by ( x+ 3)( x−
1) , we
obtain: x = A( x− 1) + B( x+
3)
Let x = 1 , then 1 = A(1− 1) + B(1+
3)
1= 4B
1
B =
4
Let x =− 3 , then − 3 = A( −3− 1) + B(
− 3+ 3)
− 3=−4A 3
A =
4
3 1
x
4 4 = +
2
x + 2x−3 x+ 3 x−1
34. Find the partial fraction decomposition:
2 2
x −x−8 x −x−8 =
2
( x+ 1)( x + 5x+ 6) ( x+ 1)( x+ 2)( x+
3)
A B C
= + +
x+ 1 x+ 2 x+
3
Multiplying both sides by ( x+ 1)( x+ 2)( x+
3) ,
we obtain:
2
x −x− 8 = A( x+ 2)( x+ 3) + B( x+ 1)( x+
3)
+ Cx ( + 1)( x+
2)
Let x =− 1,
then
2
( −1) −( −1) − 8 = A(
− 1+ 2)( − 1+ 3)
+ B(
− 1+ 1)( − 1+ 3)
+ C(
− 1+ 1)( − 1+ 2)
− 6= 2A
A =−3
Let x =− 2 , then
2
( −2) −( −2) − 8 = A(
− 2+ 2)( − 2+ 3)
+ B(
− 2 + 1)( − 2 + 3)
+ C(
− 2 + 1)( − 2 + 2)
− 2 =−B
B = 2
Let x =− 3 , then
2
( −3) −( −3) − 8 = A(
− 3+ 2)( − 3+ 3)
+ B(
− 3 + 1)( − 3 + 3)
+ C(
− 3 + 1)( − 3 + 2)
4= 2C
C = 2
2
x −x−8 −3
2 2
= + +
2
( x+ 1)( x + 5x+ 6) x+ 1 x+ 2 x+
3
850
35. Find the partial fraction decomposition:
2
x + 2x+ 3 Ax+ B Cx+ D
= +
2 2 2 2 2
( x + 4) x + 4 ( x + 4)
2 2
Multiplying both sides by ( x + 4) , we obtain:
2 2
x + 2x+ 3 = ( Ax+ B)( x + 4) + Cx+ D
2 3 2
x + 2x+ 3= Ax + Bx + 4Ax+ 4B+
Cx+ D
2 3 2
x + 2x+ 3 = Ax + Bx + (4 A+ C) x+ 4B+
D
A = 0 ; B = 1;
4A+ C = 2 4B+ D = 3
4(0) + C = 2 4(1) + D = 3
C = 2
D =−1
2
x + 2x+ 3 1 2x−1 = +
2 2 2 2 2
( x + 4) x + 4 ( x + 4)
36. Find the partial fraction decomposition:
3
x + 1 Ax + B Cx + D
= +
2 2 2 2 2
( x + 16) x + 16 ( x + 16)
2 2
Multiplying both sides by ( x + 16) , we obtain:
3 2
x + 1 = ( Ax+ B)( x + 16) + Cx+ D
3 3 2
x + 1= Ax + Bx + 16Ax+ 16B+
Cx + D
3 3 2
x + 1 = Ax + Bx + (16 A + C) x + 16B+
D
A = 1;
B = 0 ;
16A+ C = 0
16(1) + C = 0
C = −16
16B+ D = 1
16(0) + D = 1
D = 1
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3
x + 1 x − 16x+ 1
= +
2 2 2 2 2
( x + 16) x + 16 ( x + 16)
37. Find the partial fraction decomposition:
7x+ 3 7x+ 3
=
3 2
x −2x −3x
xx ( − 3)( x+
1)
A B C
= + +
x x− 3 x+
1
Multiplying both sides by xx ( − 3)( x+
1) , we
obtain:
7x+ 3 = A( x− 3)( x+ 1) + Bx( x+ 1) + Cx( x−
3)