If `2x + 5y = 4`, write another linear equation, so that lines represented by the pair are coincident.

Answer

To find another linear equation that is coincident with the given equation \(2x + 5y = 4\), we need to understand the condition for two lines to be coincident.

### Step-by-Step Solution:

1. **Identify the given equation**:
The given equation is:
\[ 2x + 5y = 4 \]

2. **Rewrite the equation in standard form**:
We can rewrite the equation in the form \(Ax + By + C = 0\):
\[ 2x + 5y - 4 = 0 \]
Here, \(A = 2\), \(B = 5\), and \(C = -4\).

3. **Use the condition for coincident lines**:
For two lines represented by equations \(A_1x + B_1y + C_1 = 0\) and \(A_2x + B_2y + C_2 = 0\) to be coincident, the following condition must hold:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]

4. **Choose a constant multiplier**:
To find another equation, we can multiply the coefficients of the original equation by any non-zero constant. Let's choose \(k\) as a constant multiplier. For example, if we choose \(k = 2\):
\[ A_2 = 2 \cdot 2 = 4, \quad B_2 = 2 \cdot 5 = 10, \quad C_2 = 2 \cdot (-4) = -8 \]

5. **Form the new equation**:
Now, substituting these values into the standard form gives us:
\[ 4x + 10y - 8 = 0 \]
This can also be written as:
\[ 4x + 10y = 8 \]

6. **Verify with another multiplier**:
We can also choose another constant, say \(k = 3\):
\[ A_2 = 3 \cdot 2 = 6, \quad B_2 = 3 \cdot 5 = 15, \quad C_2 = 3 \cdot (-4) = -12 \]
This gives us another equation:
\[ 6x + 15y - 12 = 0 \]
Which can be rewritten as:
\[ 6x + 15y = 12 \]

### Final Answer:
Thus, two equations that are coincident with the original equation \(2x + 5y = 4\) are:
1. \(4x + 10y = 8\)
2. \(6x + 15y = 12\)