If `5a+5b+20 c=t ,` then find the value of `t` for which the line `a x+b y+c-1=0` always passes through a fixed point.

Explanation

To solve the problem, we need to find the value of \( t \) for which the line \( ax + by + c - 1 = 0 \) always passes through a fixed point.

### Step-by-Step Solution:

1. **Write down the given equation**:
We have the equation \( 5a + 5b + 20c = t \). Let's denote this as Equation (1).

2. **Identify the line equation**:
The line is given by \( ax + by + c - 1 = 0 \). We can rearrange this to \( ax + by + c = 1 \).

3. **Multiply the line equation to match coefficients**:
To match the coefficients in the equation \( 5a + 5b + 20c = t \), we can multiply the entire line equation by 20:
\[ 20(ax + by + c) = 20 \cdot 1 \]
This simplifies to:
\[ 20ax + 20by + 20c = 20 \]

4. **Rearranging the equation**:
We can rearrange this to:
\[ 20ax + 20by + 20c - 20 = 0 \]
This can be rewritten as:
\[ 20ax + 20by + 20c = 20 \]

5. **Comparing coefficients**:
Now, we compare the coefficients from both equations:
- From \( 5a + 5b + 20c = t \) and \( 20ax + 20by + 20c = 20 \), we can see that for the line to always pass through a fixed point, the coefficients of \( a, b, c \) must satisfy the equation:
\[ t = 20 \]

6. **Conclusion**:
Therefore, the value of \( t \) for which the line \( ax + by + c - 1 = 0 \) always passes through a fixed point is:
\[ \boxed{20} \]