PLS HALP

Do not post A o P S homework.

Cross multiply: $3(a+2) = b(a+5)$

Simplify: $3a + 6 = ab + 5b$

Isolate the constant: $3a -ab-5b= -6$

Factor out a: $a(3-b) - 5b = -6$

Add 15 to both sides: $a(3-b) - 5b + 15 = 9$

Factor out b: $a(3-b) + 5 (3 - b) = 9$

Combine like terms: $(a+5)(3-b) = 9$

Note that $a + 5$ must equal 1, 3, or 9, so $3- b$ can equal the other factor of 9.

Thus, there are $\color{brown}\boxed{3}$ cases that work.

Note that the cases are (4, 2), (-2, 0), and (-4, -6)