Suppose \(3b-4a = 24\). Given that \(a \) and \(b\) are consecutive integers, and b

Guest Aug 8, 2022

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Aug 8, 2022

edited by Guest Aug 8, 2022

edited by Guest Aug 8, 2022

edited by Guest Aug 8, 2022

edited by Guest Aug 8, 2022

#7**0 **

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 \(\color{brown}\boxed{-28, -27}\)

BuilderBoi Aug 8, 2022

#3**0 **

Are you sure your question is formatted properly?

It just ends with ",and b"...

BuilderBoi Aug 8, 2022

#7**0 **

Best Answer

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 \(\color{brown}\boxed{-28, -27}\)

BuilderBoi Aug 8, 2022