Question

solve for v.
$v - 1\frac{4}{9} = \frac{2}{3}$
$v = \square$

Explanation:

Step1: Convert mixed number to improper fraction

First, convert \(1\frac{4}{9}\) 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 \(1\frac{4}{9}\), we have \(a = 1\), \(b = 4\), \(c = 9\). Then \(1\times9+4 = 13\), so \(1\frac{4}{9}=\frac{13}{9}\). The equation becomes \(v-\frac{13}{9}=\frac{2}{3}\).

Step2: Isolate v by adding \(\frac{13}{9}\) to both sides

To solve for \(v\), we add \(\frac{13}{9}\) to both sides of the equation. So \(v=\frac{2}{3}+\frac{13}{9}\).

Step3: Find a common denominator and add the fractions

The common denominator of 3 and 9 is 9. Convert \(\frac{2}{3}\) to a fraction with denominator 9: \(\frac{2}{3}=\frac{2\times3}{3\times3}=\frac{6}{9}\). Now add \(\frac{6}{9}\) and \(\frac{13}{9}\): \(\frac{6 + 13}{9}=\frac{19}{9}\). We can also convert this back to a mixed number: \(\frac{19}{9}=2\frac{1}{9}\), but since the problem seems to accept improper fractions or mixed numbers, and the box has a fraction format, \(\frac{19}{9}\) or \(2\frac{1}{9}\) are correct. Let's check with the steps:

Starting from \(v-\frac{13}{9}=\frac{2}{3}\), add \(\frac{13}{9}\) to both sides:

\(v=\frac{2}{3}+\frac{13}{9}\)

Find a common denominator (9):

\(\frac{2\times3}{3\times3}+\frac{13}{9}=\frac{6}{9}+\frac{13}{9}=\frac{19}{9}\) or \(2\frac{1}{9}\).

Answer:

\(\frac{19}{9}\) (or \(2\frac{1}{9}\))