Partial fraction decomposition examples with rational expressions.

Problem: Solve the given partial fraction decompositions.

We are tasked with decomposing the given rational expressions into partial fractions. Let's solve each problem step by step.

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#### 1. Decompose:
$$
\frac{9x - 7}{(3x + 4)(x - 5)} = \frac{A}{3x + 4} + \frac{B}{x - 5}
$$

##### Step 1: Write the equation in terms of a common denominator.
The right-hand side has a common denominator of $(3x + 4)(x - 5)$. Combine the fractions:
$$
\frac{A}{3x + 4} + \frac{B}{x - 5} = \frac{A(x - 5) + B(3x + 4)}{(3x + 4)(x - 5)}
$$

Equating the numerators, we get:
$$
9x - 7 = A(x - 5) + B(3x + 4)
$$

##### Step 2: Expand and collect like terms.
Expand the right-hand side:
$$
A(x - 5) + B(3x + 4) = Ax - 5A + 3Bx + 4B
$$
Combine like terms:
$$
Ax + 3Bx - 5A + 4B = (A + 3B)x + (-5A + 4B)
$$

So the equation becomes:
$$
9x - 7 = (A + 3B)x + (-5A + 4B)
$$

##### Step 3: Equate coefficients.
By comparing coefficients of $x$ and the constant terms on both sides, we get the system of equations:
$$
A + 3B = 9 \quad \text{(coefficient of } x\text{)}
$$
$$
-5A + 4B = -7 \quad \text{(constant term)}
$$

##### Step 4: Solve the system of equations.
We have:
1. $A + 3B = 9$
2. $-5A + 4B = -7$

From the first equation, solve for $A$:
$$
A = 9 - 3B
$$

Substitute $A = 9 - 3B$ into the second equation:
$$
-5(9 - 3B) + 4B = -7
$$
Simplify:
$$
-45 + 15B + 4B = -7
$$
$$
-45 + 19B = -7
$$
$$
19B = 38
$$
$$
B = 2
$$

Now substitute $B = 2$ back into $A = 9 - 3B$:
$$
A = 9 - 3(2) = 9 - 6 = 3
$$

##### Step 5: Write the partial fraction decomposition.
The values are $A = 3$ and $B = 2$. Thus, the decomposition is:
$$
\frac{9x - 7}{(3x + 4)(x - 5)} = \frac{3}{3x + 4} + \frac{2}{x - 5}
$$

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#### 2. Decompose:
$$
\frac{2x + 5}{(x - 3)^2} = \frac{A}{x - 3} + \frac{B}{(x - 3)^2}
$$

##### Step 1: Write the equation in terms of a common denominator.
The right-hand side has a common denominator of $(x - 3)^2$. Combine the fractions:
$$
\frac{A}{x - 3} + \frac{B}{(x - 3)^2} = \frac{A(x - 3) + B}{(x - 3)^2}
$$

Equating the numerators, we get:
$$
2x + 5 = A(x - 3) + B
$$

##### Step 2: Expand and collect like terms.
Expand the right-hand side:
$$
A(x - 3) + B = Ax - 3A + B
$$

So the equation becomes:
$$
2x + 5 = Ax - 3A + B
$$

##### Step 3: Equate coefficients.
By comparing coefficients of $x$ and the constant terms on both sides, we get the system of equations:
$$
A = 2 \quad \text{(coefficient of } x\text{)}
$$
$$
-3A + B = 5 \quad \text{(constant term)}
$$

##### Step 4: Solve the system of equations.
From the first equation, we have:
$$
A = 2
$$

Substitute $A = 2$ into the second equation:
$$
-3(2) + B = 5
$$
$$
-6 + B = 5
$$
$$
B = 11
$$

##### Step 5: Write the partial fraction decomposition.
The values are $A = 2$ and $B = 11$. Thus, the decomposition is:
$$
\frac{2x + 5}{(x - 3)^2} = \frac{2}{x - 3} + \frac{11}{(x - 3)^2}
$$

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#### 3. Decompose:
$$
\frac{3x + 1}{(2x + 3)(x^2 + 7x - 1)} = \frac{A}{2x + 3} + \frac{Bx + C}{x^2 + 7x - 1}
$$

##### Step 1: Write the equation in terms of a common denominator.
The right-hand side has a common denominator of $(2x + 3)(x^2 + 7x - 1)$. Combine the fractions:
$$
\frac{A}{2x + 3} + \frac{Bx + C}{x^2 + 7x - 1} = \frac{A(x^2 + 7x - 1) + (Bx + C)(2x + 3)}{(2x + 3)(x^2 + 7x - 1)}
$$

Equating the numerators, we get:
$$
3x + 1 = A(x^2 + 7x - 1) + (Bx + C)(2x + 3)
$$

##### Step 2: Expand and collect like terms.
Expand the right-hand side:
$$
A(x^2 + 7x - 1) = Ax^2 + 7Ax - A
$$
$$
(Bx + C)(2x + 3) = 2Bx^2 + 3Bx + 2Cx + 3C = 2Bx^2 + (3B + 2C)x + 3C
$$

Combine these:
$$
Ax^2 + 7Ax - A + 2Bx^2 + (3B + 2C)x + 3C = (A + 2B)x^2 + (7A + 3B + 2C)x + (-A + 3C)
$$

So the equation becomes:
$$
3x + 1 = (A + 2B)x^2 + (7A + 3B + 2C)x + (-A + 3C)
$$

##### Step 3: Equate coefficients.
By comparing coefficients of $x^2$, $x$, and the constant terms on both sides, we get the system of equations:
$$
A + 2B = 0 \quad \text{(coefficient of } x^2\text{)}
$$
$$
7A + 3B + 2C = 3 \quad \text{(coefficient of } x\text{)}
$$
$$
-A + 3C = 1 \quad \text{(constant term)}
$$

##### Step 4: Solve the system of equations.
From the first equation, solve for $A$:
$$
A = -2B
$$

Substitute $A = -2B$ into the second equation:
$$
7(-2B) + 3B + 2C = 3
$$
$$
-14B + 3B + 2C = 3
$$
$$
-11B + 2C = 3 \quad \text{(Equation 2')}
$$

Substitute $A = -2B$ into the third equation:
$$
-(-2B) + 3C = 1
$$
$$
2B + 3C = 1 \quad \text{(Equation 3')}
$$

Now solve the system of two equations:
1. $-11B + 2C = 3$
2. $2B + 3C = 1$

Multiply the second equation by 2 to eliminate $C$:
$$
4B + 6C = 2
$$

Multiply the first equation by 3:
$$
-33B + 6C = 9
$$

Subtract the first modified equation from the second:
$$
(4B + 6C) - (-33B + 6C) = 2 - 9
$$
$$
4B + 33B = -7
$$
$$
37B = -7
$$
$$
B = -\frac{7}{37}
$$

Substitute $B = -\frac{7}{37}$ into $2B + 3C = 1$:
$$
2\left(-\frac{7}{37}\right) + 3C = 1
$$
$$
-\frac{14}{37} + 3C = 1
$$
$$
3C = 1 + \frac{14}{37}
$$
$$
3C = \frac{37}{37} + \frac{14}{37} = \frac{51}{37}
$$
$$
C = \frac{51}{37} \cdot \frac{1}{3} = \frac{17}{37}
$$

Substitute $B = -\frac{7}{37}$ into $A = -2B$:
$$
A = -2\left(-\frac{7}{37}\right) = \frac{14}{37}
$$

##### Step 5: Write the partial fraction decomposition.
The values are $A = \frac{14}{37}$, $B = -\frac{7}{37}$, and $C = \frac{17}{37}$. Thus, the decomposition is:
$$
\frac{3x + 1}{(2x + 3)(x^2 + 7x - 1)} = \frac{\frac{14}{37}}{2x + 3} + \frac{-\frac{7}{37}x + \frac{17}{37}}{x^2 + 7x - 1}
$$
Simplify:
$$
\frac{3x + 1}{(2x + 3)(x^2 + 7x - 1)} = \frac{14}{37(2x + 3)} + \frac{-7x + 17}{37(x^2 + 7x - 1)}
$$

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Final Answers:

1. $\boxed{\frac{3}{3x + 4} + \frac{2}{x - 5}}$
2. $\boxed{\frac{2}{x - 3} + \frac{11}{(x - 3)^2}}$
3. $\boxed{\frac{14}{37(2x + 3)} + \frac{-7x + 17}{37(x^2 + 7x - 1)}}$