If 2^p + 3^q = 17 and 2^{p + 2} - 3^{q + 1} = 5, then, find the value of p and q?

Apply the law of Fractional Exponents and Laws of Exponents
(a^m) x (aⁿ) = a^{m + n}
a^-m = 1/a^m
p⁰ = 1
and if p^X = p^Y then X will be equal to Y. means X = Y;
(a^m)ⁿ = a^{m x n}
(a^m) x (aⁿ) = a^{m + n}
a^m ÷ aⁿ = a^{m ? n}
Or
a^m/aⁿ = a^{m ? n}

Given equation is
2^p + 3^q = 17 ----------------(i)
2^{p + 2} - 3^{q + 1} = 5
? 2^p x 2² - 3^q x 3 = 5
? 2^p x 4 - 3q x 3 = 5
? 4 x 2^p - 3 x 3q = 5---------------(ii)
On multiplying Eq. (i) by 3 and adding it, with Eq. (ii), we get

3 x 2^p + 3 x 3^q = 51
4 x 2^p - 3 x 3^q = 5
_____________________________
7 x 2^p = 56

? 7 x 2^p = 7 x 8
? 2^p = 8
? 2^p = 2³
? p = 3

On substituting the value of p in Eq. (i) . we get
2³ + 3^q = 17
? 8 + 3^q = 17
? 3^q = 17 - 8
? 3^q = 9
? 3^q = 3²
? q = 2