Suppose \(3b-4a = 24\). Given that \(a \) and \(b\) are consecutive integers, and b
There are 2 cases to consider: \(a = b + 1 \) and \(b = a + 1\)
So, we have 2 systems:
\(3a - 4b = 24\)
\(a = b + 1\)
AND
\(3a - 4b = 24\)
\(b = a + 1\)
The solutions for the two systems, respectively in the form of \((a,b)\), are \(\color{brown}\boxed{-20, -21}\) and \(\color{brown}\boxed{-28, -27}\)
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There are 2 cases to consider: \(a = b + 1 \) and \(b = a + 1\)
So, we have 2 systems:
\(3a - 4b = 24\)
\(a = b + 1\)
AND
\(3a - 4b = 24\)
\(b = a + 1\)
The solutions for the two systems, respectively in the form of \((a,b)\), are \(\color{brown}\boxed{-20, -21}\) and \(\color{brown}\boxed{-28, -27}\)