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?

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.