Use the graphical method to find the value of k if
(i) (k,-3) lies on the straight line
`2x+3y=1`
(ii) (5,k-2) lies on the straight line
`x-2y+1=0`

Answer

To solve the problem using the graphical method, we need to find the value of \( k \) for the two given conditions. Let's break it down step by step.

### Step 1: Solve for \( k \) in the first equation

**Given:** The point \( (k, -3) \) lies on the line defined by the equation \( 2x + 3y = 1 \).

1. Substitute \( y = -3 \) into the equation:
\[ 2x + 3(-3) = 1 \]
Simplifying this gives:
\[ 2x - 9 = 1 \]
\[ 2x = 1 + 9 \]
\[ 2x = 10 \]
\[ x = 5 \]

2. Since \( x = k \), we find that:
\[ k = 5 \]

### Step 2: Solve for \( k \) in the second equation

**Given:** The point \( (5, k-2) \) lies on the line defined by the equation \( x - 2y + 1 = 0 \).

1. Substitute \( x = 5 \) into the equation:
\[ 5 - 2y + 1 = 0 \]
Simplifying this gives:
\[ 6 - 2y = 0 \]
\[ 2y = 6 \]
\[ y = 3 \]

2. Since \( y = k - 2 \), we have:
\[ k - 2 = 3 \]
\[ k = 3 + 2 \]
\[ k = 5 \]

### Conclusion

From both parts of the problem, we find that:
\[ \boxed{5} \]