Find the values of k for which the following equations have an infinite number of solutions: 2x + 3y =7 , (k-1)x+(k+2)y=3k

Answer

To find the values of \( k \) for which the equations \( 2x + 3y = 7 \) and \( (k-1)x + (k+2)y = 3k \) have an infinite number of solutions, we follow these steps:

### Step 1: Write the equations in standard form
We can rewrite the given equations in the standard form \( ax + by + c = 0 \).

1. The first equation:
\[ 2x + 3y - 7 = 0 \]
Here, \( a_1 = 2 \), \( b_1 = 3 \), and \( c_1 = -7 \).

2. The second equation:
\[ (k-1)x + (k+2)y - 3k = 0 \]
Here, \( a_2 = k-1 \), \( b_2 = k+2 \), and \( c_2 = -3k \).

### Step 2: Set up the condition for infinite solutions
For the two equations to have an infinite number of solutions, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

### Step 3: Write the ratios
Substituting the values we have:
\[ \frac{2}{k-1} = \frac{3}{k+2} = \frac{-7}{-3k} \]

### Step 4: Solve the first ratio
Let's solve the first part:
\[ \frac{2}{k-1} = \frac{3}{k+2} \]
Cross-multiplying gives:
\[ 2(k + 2) = 3(k - 1) \]
Expanding both sides:
\[ 2k + 4 = 3k - 3 \]
Rearranging gives:
\[ 4 + 3 = 3k - 2k \]
\[ 7 = k \]

### Step 5: Solve the second ratio
Now, let's solve the second part:
\[ \frac{3}{k+2} = \frac{7}{3k} \]
Cross-multiplying gives:
\[ 3 \cdot 3k = 7(k + 2) \]
Expanding both sides:
\[ 9k = 7k + 14 \]
Rearranging gives:
\[ 9k - 7k = 14 \]
\[ 2k = 14 \]
Dividing by 2 gives:
\[ k = 7 \]

### Conclusion
Both cases yield \( k = 7 \). Therefore, the value of \( k \) for which the equations have an infinite number of solutions is:
\[ \boxed{7} \]