x:(2/3+1/9) =9/35
To solve for \( x \) in the equation
\[ x : \left( \frac{2}{3} + \frac{1}{9} \right) = \frac{9}{35}, \]
we first need to simplify the expression on the left-hand side.
Add the fractions \(\frac{2}{3}\) and \(\frac{1}{9}\):
To add these fractions, we need a common denominator. The least common multiple of 3 and 9 is 9.
- Convert \(\frac{2}{3}\) to a fraction with a denominator of 9:
\[ \frac{2}{3} = \frac{2 \times 3}{3 \times 3} = \frac{6}{9}. \]
- Now we can add:
\[ \frac{6}{9} + \frac{1}{9} = \frac{6 + 1}{9} = \frac{7}{9}. \]
Rewrite the equation:
Now, substituting this back into the equation gives:
\[ x : \frac{7}{9} = \frac{9}{35}. \]
This can be understood as:
\[ \frac{x}{\frac{7}{9}} = \frac{9}{35}. \]
Cross-multiply to solve for \( x \):
The equation becomes:
\[ x \cdot 35 = \frac{9}{35} \cdot 7. \]
Calculate \( \frac{9 \cdot 7}{35} \):
\[ 9 \cdot 7 = 63, \] so,
\[ x \cdot 35 = \frac{63}{35}. \]
Multiply each side by \(\frac{1}{35}\):
\[ x = \frac{63}{35} \cdot \frac{1}{35} = \frac{63}{35} \times \frac{1}{35}. \]
Simplifying \(\frac{63}{35}\):
The GCD (Greatest Common Divisor) of 63 and 35 is 7:
\[ \frac{63 \div 7}{35 \div 7} = \frac{9}{5}. \]
Final Solution:
Therefore, the value of \( x \) is:
\[ x = \frac{9}{5}. \]
In conclusion:
\[ x = \frac{9}{5}. \]
