Question

number - fractions - add unlike mixed numbers
find the sum. simplify your answer.
$4 \frac{18}{48} + 2 \frac{3}{12} = \square \frac{\square}{\square}$

Explanation:

Step1: Simplify the fractions

First, simplify \(\frac{18}{48}\) by dividing numerator and denominator by 6: \(\frac{18\div6}{48\div6}=\frac{3}{8}\). Then, simplify \(\frac{3}{12}\) by dividing numerator and denominator by 3: \(\frac{3\div3}{12\div3}=\frac{1}{4}\). Now the problem becomes \(4\frac{3}{8}+2\frac{1}{4}\).

Step2: Convert mixed numbers to improper fractions (or add whole numbers and fractions separately)

Add the whole numbers: \(4 + 2 = 6\). Add the fractions: \(\frac{3}{8}+\frac{1}{4}\). Convert \(\frac{1}{4}\) to eighths: \(\frac{1\times2}{4\times2}=\frac{2}{8}\). Then \(\frac{3}{8}+\frac{2}{8}=\frac{5}{8}\).

Step3: Combine whole number and fraction

Combine the results: \(6+\frac{5}{8}=6\frac{5}{8}\). Wait, but let's check again by converting to improper fractions. \(4\frac{3}{8}=\frac{4\times8 + 3}{8}=\frac{35}{8}\), \(2\frac{1}{4}=\frac{2\times4+1}{4}=\frac{9}{4}=\frac{18}{8}\). Then \(\frac{35}{8}+\frac{18}{8}=\frac{53}{8}=6\frac{5}{8}\). Wait, but originally the first fraction was \(4\frac{18}{48}\), let's simplify \(\frac{18}{48}\) to \(\frac{3}{8}\) (since 18÷6=3, 48÷6=8) and the second fraction \(\frac{3}{12}=\frac{1}{4}=\frac{2}{8}\). So adding fractions: \(\frac{18}{48}+\frac{3}{12}=\frac{18}{48}+\frac{12}{48}=\frac{30}{48}=\frac{5}{8}\). Then whole numbers: 4 + 2 = 6. So total is \(6\frac{5}{8}\). Wait, but let's do it step by step with the original fractions.

First, simplify \(\frac{18}{48}\): GCD of 18 and 48 is 6, so \(\frac{18\div6}{48\div6}=\frac{3}{8}\). Simplify \(\frac{3}{12}\): GCD of 3 and 12 is 3, so \(\frac{3\div3}{12\div3}=\frac{1}{4}\). Now, add the mixed numbers:

Whole parts: \(4 + 2 = 6\)

Fraction parts: \(\frac{3}{8}+\frac{1}{4}\). To add these, common denominator is 8. \(\frac{1}{4}=\frac{2}{8}\), so \(\frac{3}{8}+\frac{2}{8}=\frac{5}{8}\)

So total is \(6+\frac{5}{8}=6\frac{5}{8}\). Wait, but let's check the original calculation with the unsimplified fractions. \(4\frac{18}{48}+2\frac{3}{12}\). Convert \(2\frac{3}{12}\) to forty-eighths: \(\frac{3}{12}=\frac{12}{48}\), so \(2\frac{12}{48}\). Now add \(4\frac{18}{48}+2\frac{12}{48}=(4 + 2)+(\frac{18}{48}+\frac{12}{48})=6+\frac{30}{48}=6+\frac{5}{8}=6\frac{5}{8}\) (since \(\frac{30}{48}=\frac{5}{8}\) after dividing numerator and denominator by 6).

Answer:

\(6\frac{5}{8}\)