Question

2\dfrac{1}{9} - \dfrac{2}{4} =

Explanation:

Step1: Convert mixed numbers to improper fractions

First, convert \( 2\frac{1}{9} \) to an improper fraction. The formula for converting a mixed number \( a\frac{b}{c} \) to an improper fraction is \( \frac{a\times c + b}{c} \). So for \( 2\frac{1}{9} \), we have \( a = 2 \), \( b = 1 \), \( c = 9 \). Then \( 2\frac{1}{9}=\frac{2\times9 + 1}{9}=\frac{18 + 1}{9}=\frac{19}{9} \).

Next, convert \( \frac{2}{4} \) (wait, no, the second term is \( \frac{2}{4} \)? Wait, the original problem is \( 2\frac{1}{9}-\frac{2}{4} \)? Wait, no, the second term is \( \frac{2}{4} \)? Wait, maybe it's a typo, but assuming it's \( 2\frac{1}{9}-\frac{2}{4} \), but let's check. Wait, maybe the second term is a mixed number? Wait, the image shows \( 2\frac{1}{9}-\frac{2}{4} \)? Wait, no, the second term is written as \( \frac{2}{4} \) with a 2 above? Wait, maybe it's \( 2\frac{1}{9}-\frac{2}{4} \), but let's proceed. Wait, maybe the second term is \( \frac{2}{4} \), but let's simplify.

Wait, first, let's correct: the problem is \( 2\frac{1}{9}-\frac{2}{4} \)? Wait, no, maybe the second term is \( \frac{2}{4} \), but let's convert \( 2\frac{1}{9} \) to improper fraction: \( \frac{19}{9} \), and \( \frac{2}{4}=\frac{1}{2} \). Now, find a common denominator for 9 and 2, which is 18.

Step2: Find common denominator and subtract

Convert \( \frac{19}{9} \) to eighteenths: \( \frac{19}{9}=\frac{19\times2}{9\times2}=\frac{38}{18} \).

Convert \( \frac{1}{2} \) to eighteenths: \( \frac{1}{2}=\frac{1\times9}{2\times9}=\frac{9}{18} \).

Now subtract: \( \frac{38}{18}-\frac{9}{18}=\frac{38 - 9}{18}=\frac{29}{18} \).

Wait, but maybe the second term is a mixed number? Wait, the image shows \( 2\frac{1}{9}-\frac{2}{4} \)? No, maybe the second term is \( \frac{2}{4} \), but let's check again. Wait, maybe the original problem is \( 2\frac{1}{9}-\frac{2}{4} \), but let's simplify \( \frac{2}{4}=\frac{1}{2} \).

Alternatively, maybe the second term is \( \frac{2}{4} \) as a fraction. Wait, perhaps I misread. Let's re-express:

Original problem: \( 2\frac{1}{9}-\frac{2}{4} \)

Step1: Convert \( 2\frac{1}{9} \) to improper fraction: \( \frac{19}{9} \)

Step2: Simplify \( \frac{2}{4}=\frac{1}{2} \)

Step3: Find common denominator (9 and 2, LCM is 18)

\( \frac{19}{9}=\frac{38}{18} \), \( \frac{1}{2}=\frac{9}{18} \)

Step4: Subtract: \( \frac{38}{18}-\frac{9}{18}=\frac{29}{18}=1\frac{11}{18} \)

Wait, but maybe the second term is \( \frac{2}{4} \) as a mixed number? No, \( \frac{2}{4} \) is a proper fraction. Wait, maybe the original problem is \( 2\frac{1}{9}-\frac{2}{4} \), so let's do that.

Answer:

\( 1\frac{11}{18} \) (or \( \frac{29}{18} \))