To add \(-\frac{12}{7}\) and \(3 \frac{1}{4}\), we first need to convert the mixed number \(3 \frac{1}{4}\) into an improper fraction.
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Convert \(3 \frac{1}{4}\) to an improper fraction: \[ 3 \frac{1}{4} = 3 \cdot 4 + 1 = 12 + 1 = 13 \quad \text{(so it becomes } \frac{13}{4}\text{)} \]
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Now we need to add \(-\frac{12}{7}\) and \(\frac{13}{4}\). Before doing that, we need a common denominator. The denominators are 7 and 4. The least common multiple of 7 and 4 is 28.
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Convert both fractions to have the denominator of 28:
- For \(-\frac{12}{7}\): \[ -\frac{12}{7} = -\frac{12 \cdot 4}{7 \cdot 4} = -\frac{48}{28} \]
- For \(\frac{13}{4}\): \[ \frac{13}{4} = \frac{13 \cdot 7}{4 \cdot 7} = \frac{91}{28} \]
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Now add the two fractions: \[ -\frac{48}{28} + \frac{91}{28} = \frac{91 - 48}{28} = \frac{43}{28} \]
The result \(\frac{43}{28}\) is already in reduced form. Therefore, the answer is:
\[ \boxed{\frac{43}{28}} \]