4k = 25 => k = 25/4 => 6< k < 7 15q = 160 + 24 => 15q = 184 => q = 184/15 => 12< q < 13
The solution for k in the equation 4 k − 13 = 12 lies between the integers 6 and 7, while the solution for q in the equation 24.1 = 15 q − 32.5 lies between 3 and 4. Thus, the two sets of consecutive integers are (6, 7) for k and (3, 4) for q.
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