To solve this equation, we first need to understand the notation being used. The notation seems to represent the numbers in a base, with parentheses indicating a repeating decimal.
The equation appears to be:
[ 0.121212... : 0.166666... = x : 0.33333... ]
Let's convert these repeating decimals into fractions:
Convert (0.121212...) to a fraction:
Let (y = 0.121212...).
Then, (100y = 12.121212...).
Subtracting the original (y):
[
100y - y = 12.121212... - 0.121212...
]
[
99y = 12 \implies y = \frac{12}{99} = \frac{4}{33}
]
Convert (0.166666...) to a fraction:
Let (z = 0.166666...).
Then, (10z = 1.66666...).
Subtracting gives:
[
10z - z = 1.66666... - 0.16666...
]
[
9z = 1 \implies z = \frac{1}{9}
]
Convert (0.33333...) to a fraction:
(0.33333... = \frac{1}{3}).
Now substituting these fractions back into our equation:
[
\frac{\frac{4}{33}}{\frac{1}{9}} = \frac{x}{\frac{1}{3}}
]
To simplify the left side:
[
\frac{4}{33} \div \frac{1}{9} = \frac{4}{33} \times 9 = \frac{36}{33} = \frac{12}{11}
]
So we have:
[
\frac{12}{11} = \frac{x}{\frac{1}{3}}
]
Now cross-multiplying to solve for (x):
[
12 \cdot \frac{1}{3} = 11x \implies 4 = 11x \implies x = \frac{4}{11}
]
Thus, the solution to the equation is:
[
x = \frac{4}{11}
]
