To subtract \(2\) wholes and \(7\) twelfths from \(4\), we can express the whole number \(4\) as a fraction with a denominator of \(12\).
First, convert \(4\) into twelfths:
\[ 4 = \frac{4 \times 12}{1 \times 12} = \frac{48}{12} \]
Next, we want to subtract \(2\) wholes and \(7\) twelfths. In fraction form, \(2\) wholes is:
\[ 2 = \frac{2 \times 12}{1 \times 12} = \frac{24}{12} \]
Now, \(2\) wholes and \(7\) twelfths can be written as:
\[ 2 + \frac{7}{12} = \frac{24}{12} + \frac{7}{12} = \frac{31}{12} \]
Now we can perform the subtraction:
\[ 4 - (2 + \frac{7}{12}) = \frac{48}{12} - \frac{31}{12} \]
Subtract the fractions:
\[ \frac{48 - 31}{12} = \frac{17}{12} \]
So, \(4\) minus \(2\) wholes and \(7\) twelfths equals:
\[ \frac{17}{12} \]
This can also be expressed as:
\[ 1 \frac{5}{12} \]
Thus, \(4 - 2 \text{ wholes and } 7 \text{ twelfths} = \frac{17}{12}\) or \(1 \frac{5}{12}\).