$\begin{array}{l}\frac{1 \cdot 3}{4 \cdot 3}+\frac{1 \cdot 2}{6 \cdot 2}= \\ \frac{3}{12}+\frac{2}{12}=\frac{5}{12}\end{array}$
We are asked to verify the equality of the mathematical expression:
\[\frac{1 \cdot 3}{4 \cdot 3} + \frac{1 \cdot 2}{6 \cdot 2} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}.\]
This expression involves adding two fractions by first converting them to a common denominator of 12. We will analyze each part of this equality chain to confirm its accuracy.
First, we evaluate the first term: \(\frac{1 \cdot 3}{4 \cdot 3}\).
Multiplying the numerator and the denominator, we get:
\[\frac{1 \cdot 3}{4 \cdot 3} = \frac{3}{12}.\]
This simplifies to \(\frac{1}{4}\) because both the numerator and denominator are divisible by 3. Thus, the first part of the equality \(\frac{1 \cdot 3}{4 \cdot 3} = \frac{3}{12}\) is correct.
Next, we evaluate the second term: \(\frac{1 \cdot 2}{6 \cdot 2}\).
Multiplying the numerator and the denominator, we get:
\[\frac{1 \cdot 2}{6 \cdot 2} = \frac{2}{12}.\]
This simplifies to \(\frac{1}{6}\) because both the numerator and denominator are divisible by 2. Thus, the second part of the equality \(\frac{1 \cdot 2}{6 \cdot 2} = \frac{2}{12}\) is correct.
Now, we compute the sum of these fractions using the common denominator of 12:
\[\frac{3}{12} + \frac{2}{12} = \frac{3+2}{12} = \frac{5}{12}.\]
Since both components of the sum are correct, the entire expression \(\frac{3}{12} + \frac{2}{12} = \frac{5}{12}\) is correct. Therefore, the original equality holds true.
Adding fractions is essential in daily life, such as in cooking where you might need to combine different measurements. For instance, if a recipe calls for \(\frac{1}{4}\) cup of flour and \(\frac{1}{6}\) cup of sugar, and you want to know the total volume of these dry ingredients, you would perform exactly this type of calculation: \(\frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}\) of a cup. This simple arithmetic helps ensure your measurements are precise and your culinary results turn out perfectly every time.