The problem asks: What number should be subtracted from the sum of $2\frac{1}{6}$ and $2\frac{7}{12}$ to give $3\frac{1}{4}$?

1. Problem Description

The problem asks: What number should be subtracted from the sum of $2\frac{1}{6}$ and $2\frac{7}{12}$ to give $3\frac{1}{4}$?

2. Solution Steps

Let $x$ be the number that should be subtracted.

We can set up the equation:

$(2\frac{1}{6} + 2\frac{7}{12}) - x = 3\frac{1}{4}$

First, convert the mixed numbers to improper fractions:

$2\frac{1}{6} = \frac{2 \times 6 + 1}{6} = \frac{13}{6}$

$2\frac{7}{12} = \frac{2 \times 12 + 7}{12} = \frac{31}{12}$

$3\frac{1}{4} = \frac{3 \times 4 + 1}{4} = \frac{13}{4}$

Now, substitute the improper fractions into the equation:

$(\frac{13}{6} + \frac{31}{12}) - x = \frac{13}{4}$

Find a common denominator for $\frac{13}{6}$ and $\frac{31}{12}$, which is

1

2. $\frac{13}{6} = \frac{13 \times 2}{6 \times 2} = \frac{26}{12}$

So, the equation becomes:

$(\frac{26}{12} + \frac{31}{12}) - x = \frac{13}{4}$

$\frac{26+31}{12} - x = \frac{13}{4}$

$\frac{57}{12} - x = \frac{13}{4}$

Now, find a common denominator for $\frac{57}{12}$ and $\frac{13}{4}$, which is

1

2. $\frac{13}{4} = \frac{13 \times 3}{4 \times 3} = \frac{39}{12}$

The equation is now:

$\frac{57}{12} - x = \frac{39}{12}$

$x = \frac{57}{12} - \frac{39}{12}$

$x = \frac{57 - 39}{12}$

$x = \frac{18}{12}$

Simplify the fraction:

$x = \frac{6 \times 3}{6 \times 2} = \frac{3}{2}$

Convert the improper fraction to a mixed number:

$x = 1\frac{1}{2}$

3. Final Answer

$1\frac{1}{2}$