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We are given the equation:

\[
p(x + 2)(x + 1) + q(x - 3)(x - 1) + r(x - 2)(x - 6) = 2x^2 + 7x - 13
\]

We need to find the values of \( p \), \( q \), and \( r \).

### Step 1: Expand the terms on the left-hand side

#### Expand \( p(x + 2)(x + 1) \):
\[
p(x + 2)(x + 1) = p(x^2 + x + 2x + 2) = p(x^2 + 3x + 2) = p \cdot x^2 + 3p \cdot x + 2p
\]

#### Expand \( q(x - 3)(x - 1) \):
\[
q(x - 3)(x - 1) = q(x^2 - x - 3x + 3) = q(x^2 - 4x + 3) = q \cdot x^2 - 4q \cdot x + 3q
\]

#### Expand \( r(x - 2)(x - 6) \):
\[
r(x - 2)(x - 6) = r(x^2 - 6x - 2x + 12) = r(x^2 - 8x + 12) = r \cdot x^2 - 8r \cdot x + 12r
\]

### Step 2: Combine all expanded terms

Now, combine all terms on the left-hand side:
\[
p \cdot x^2 + 3p \cdot x + 2p + q \cdot x^2 - 4q \cdot x + 3q + r \cdot x^2 - 8r \cdot x + 12r
\]
Group the like terms (powers of \( x \)):
\[
(p + q + r) \cdot x^2 + (3p - 4q - 8r) \cdot x + (2p + 3q + 12r)
\]

### Step 3: Set up equations by comparing coefficients

We now compare the coefficients of \( x^2 \), \( x \), and the constant terms with the right-hand side \( 2x^2 + 7x - 13 \).

1. Coefficient of \( x^2 \): \( p + q + r = 2 \)
2. Coefficient of \( x \): \( 3p - 4q - 8r = 7 \)
3. Constant term: \( 2p + 3q + 12r = -13 \)

### Step 4: Solve the system of equations

The system of equations is:

1. \( p + q + r = 2 \)
2. \( 3p - 4q - 8r = 7 \)
3. \( 2p + 3q + 12r = -13 \)

Let me solve this system of equations.

From equation (1), solve for \( p \):

\[
p = 2 - q - r
\]

Substitute this expression for \( p \) into equations (2) and (3):

For equation (2):
\[
3(2 - q - r) - 4q - 8r = 7
\]
\[
6 - 3q - 3r - 4q - 8r = 7
\]
\[
6 - 7q - 11r = 7
\]
\[
-7q - 11r = 1 \quad \text{(Equation 4)}
\]

For equation (3):
\[
2(2 - q - r) + 3q + 12r = -13
\]
\[
4 - 2q - 2r + 3q + 12r = -13
\]
\[
4 + q + 10r = -13
\]
\[
q + 10r = -17 \quad \text{(Equation 5)}
\]

### Step 5: Solve the two-variable system

We now have the system:

1. \( -7q - 11r = 1 \) (Equation 4)
2. \( q + 10r = -17 \) (Equation 5)

Solve for \( q \) in terms of \( r \) from Equation (5):
\[
q = -17 - 10r
\]

Substitute this into Equation (4):
\[
-7(-17 - 10r) - 11r = 1
\]
\[
119 + 70r - 11r = 1
\]
\[
119 + 59r = 1
\]
\[
59r = 1 - 119
\]
\[
59r = -118
\]
\[
r = -2
\]

### Step 6: Solve for \( q \) and \( p \)

Substitute \( r = -2 \) into Equation (5):
\[
q + 10(-2) = -17
\]
\[
q - 20 = -17
\]
\[
q = 3
\]

Now, substitute \( q = 3 \) and \( r = -2 \) into the expression for \( p \):
\[
p = 2 - 3 - (-2) = 2 - 3 + 2 = 1
\]

### Final Solution:
\[
p = 1, \quad q = 3, \quad r = -2
\]

Would you like a more detailed breakdown or have any questions?

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Here are some related questions for practice:

1. What happens if we change the equation on the right-hand side?
2. How would you approach solving a system with more unknowns?
3. Can you find the discriminant of the quadratic expression for \( p = 2 \)?
4. How does the solution change if the coefficients on the right-hand side are different?
5. What method can we use to simplify solving this system?

**Tip**: When solving systems of equations, substitution is often effective when one variable is easy to isolate.

Given that p(x + 2)(x + 1) + q(x - 3)(x - 1) + r(x - 2)(x - 6) = 2x^2 + 7x - 13 for all values of x, find the values of p, q, and r.

This problem involves expanding polynomials and comparing coefficients to find the values of p, q, and r. Learn how to solve the system of equations step-by-step, using algebraic techniques, suitable for high school students.

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