`{:(8x + 5y = 9),(kx + 10y = 8):}`

Answer

To find the value of \( k \) in the equations \( 8x + 5y = 9 \) and \( kx + 10y = 8 \), we will use the concept of no solution for a system of linear equations.

### Step 1: Rewrite the equations in standard form
We can rewrite the given equations in the standard form \( Ax + By + C = 0 \).

1. For the first equation:
\[ 8x + 5y - 9 = 0 \]
This can be labeled as Equation 1.

2. For the second equation:
\[ kx + 10y - 8 = 0 \]
This can be labeled as Equation 2.

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For Equation 1: \( a_1 = 8 \), \( b_1 = 5 \), \( c_1 = -9 \)
- For Equation 2: \( a_2 = k \), \( b_2 = 10 \), \( c_2 = -8 \)

### Step 3: Apply the condition for no solution
For the system of equations to have no solution, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

Substituting the values we have:
\[ \frac{8}{k} = \frac{5}{10} \]
And we also need:
\[ \frac{8}{k} \neq \frac{-9}{-8} \]

### Step 4: Simplify the ratios
First, simplify \( \frac{5}{10} \):
\[ \frac{5}{10} = \frac{1}{2} \]

Now we can set up the equation:
\[ \frac{8}{k} = \frac{1}{2} \]

### Step 5: Cross-multiply to solve for \( k \)
Cross-multiplying gives:
\[ 8 \cdot 2 = 1 \cdot k \]
\[ 16 = k \]

### Step 6: Verify the second condition
Now, we need to check that:
\[ \frac{8}{k} \neq \frac{9}{8} \]
Substituting \( k = 16 \):
\[ \frac{8}{16} = \frac{1}{2} \quad \text{and} \quad \frac{9}{8} \text{ is not equal to } \frac{1}{2} \]
Thus, the condition holds.

### Final Answer
The value of \( k \) is \( 16 \).

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