if x=0.ababababab.... and y=0.bababababbaba.... such that x+y=0.cccccccc..... where a,b,c are single digit numbers then possible values of x can be

Let's analyze the given repeating decimal equations and find possible values of \( x \).

### Step 1: Express \( x \) and \( y \) as Fractions
Given:
- \( x = 0.ababababab... \) (repeating "ab")
- \( y = 0.bababababa... \) (repeating "ba")
- \( x + y = 0.ccccccc... \) (repeating "c")

Since \( x \) and \( y \) are repeating decimals, we express them as fractions.

#### Convert \( x = 0.\overline{ab} \) into Fraction
Let \( x = 0.\overline{ab} \), where "ab" represents two digits (for example, if \( a = 1 \) and \( b = 2 \), then \( x = 0.121212... \)).

Using the formula for repeating decimals:

\( x = \frac{\text{repeating part}}{\text{same number of 9s}} \)

\( x = \frac{ab}{99} \)

Since "ab" is a two-digit number, it means \( x = \frac{10a + b}{99} \).

#### Convert \( y = 0.\overline{ba} \) into Fraction
Similarly, for \( y = 0.\overline{ba} \),

\( y = \frac{ba}{99} = \frac{10b + a}{99} \).

### Step 2: Sum of \( x \) and \( y \)
Given that \( x + y = 0.\overline{c} \), which is a repeating decimal,

\( x + y = \frac{10a + b}{99} + \frac{10b + a}{99} \)

\( x + y = \frac{(10a + b) + (10b + a)}{99} \)

\( x + y = \frac{10a + b + 10b + a}{99} \)

\( x + y = \frac{11a + 11b}{99} \)

\( x + y = \frac{11(a + b)}{99} \)

\( x + y = \frac{a + b}{9} \)

Since the sum is given as \( 0.\overline{c} \), we also know that:

\( 0.\overline{c} = \frac{c}{9} \)

Thus,

\( \frac{a + b}{9} = \frac{c}{9} \)

So, \( c = a + b \).

### Step 3: Find Possible Values of \( x \)
Since \( a \) and \( b \) are single digits (0 to 9), \( c = a + b \) must also be a single digit. This means:

\( 0 \leq a + b \leq 9 \).

For each valid pair \( (a, b) \), the possible values of \( x \) are:

\( x = \frac{10a + b}{99} \).

Some valid values:

1. If \( a = 0, b = 0 \), then \( x = \frac{00}{99} = 0 \).
2. If \( a = 1, b = 2 \), then \( x = \frac{10(1) + 2}{99} = \frac{12}{99} = \frac{4}{33} \).
3. If \( a = 2, b = 3 \), then \( x = \frac{10(2) + 3}{99} = \frac{23}{99} \).
4. If \( a = 3, b = 4 \), then \( x = \frac{34}{99} \).
5. If \( a = 4, b = 5 \), then \( x = \frac{45}{99} = \frac{5}{11} \).
6. If \( a = 5, b = 4 \), then \( x = \frac{54}{99} = \frac{6}{11} \).
7. If \( a = 6, b = 3 \), then \( x = \frac{63}{99} = \frac{7}{11} \).
8. If \( a = 7, b = 2 \), then \( x = \frac{72}{99} = \frac{8}{11} \).
9. If \( a = 8, b = 1 \), then \( x = \frac{81}{99} = \frac{9}{11} \).
10. If \( a = 9, b = 0 \), then \( x = \frac{90}{99} = \frac{10}{11} \).

### Final Answer:
The possible values of \( x \) are:

\( 0, \frac{4}{33}, \frac{23}{99}, \frac{34}{99}, \frac{5}{11}, \frac{6}{11}, \frac{7}{11}, \frac{8}{11}, \frac{9}{11}, \frac{10}{11} \).