Exponential equations system (repaired exponents): If 2^p + 3^q = 17 and 2^(p+2) − 3^(q+1) = 5, find ordered pair (p, q).

Introduction / Context:
The database text omits exponent markers on the second equation. Using the Recovery-First Policy, we repair it to a standard pair of exponential equations. Substitution via linearization trick (set A = 2^p, B = 3^q) reduces the system to linear equations in A and B.

Given Data / Assumptions:

• 2^p + 3^q = 17
• 2^(p+2) − 3^(q+1) = 5
• 2^(p+2) = 4·2^p and 3^(q+1) = 3·3^q

Concept / Approach:
Let A = 2^p and B = 3^q. Then equations become A + B = 17 and 4A − 3B = 5. Solve the linear system for A and B, then back-solve for p and q by recognizing small powers of 2 and 3.

Step-by-Step Solution:
A + B = 17 … (1)4A − 3B = 5 … (2)From (1): B = 17 − APlug in (2): 4A − 3(17 − A) = 5 ⇒ 4A − 51 + 3A = 5 ⇒ 7A = 56 ⇒ A = 8Thus 2^p = 8 ⇒ p = 3; and B = 17 − 8 = 9 = 3^2 ⇒ q = 2

Verification / Alternative check:
Check both equations: 2^3 + 3^2 = 8 + 9 = 17; 2^5 − 3^3 = 32 − 27 = 5, both true.

Why Other Options Are Wrong:
They do not solve the linear system correctly or do not match integer powers of 2 and 3.

Common Pitfalls:
Forgetting 2^(p+2) = 4·2^p and 3^(q+1) = 3·3^q; mixing addition and subtraction signs.

Final Answer:
3, 2