For what value of k, will the following pair of equations have infinitely many solutions:
`2x+ 3y=7 and (k+2)x-3 (1-k) y= 5k +1`

Answer

To find the value of \( k \) for which the given pair of equations has infinitely many solutions, we need to set the equations in the standard form and apply the condition for infinitely many solutions.

The given equations are:
1. \( 2x + 3y = 7 \)
2. \( (k + 2)x - 3(1 - k)y = 5k + 1 \)

### Step 1: Rewrite the second equation in standard form
The second equation can be rewritten as:
\[ (k + 2)x - (3 - 3k)y = 5k + 1 \]
This can be expressed as:
\[ (k + 2)x + (3k - 3)y - (5k + 1) = 0 \]

### Step 2: Identify coefficients
From the first equation \( 2x + 3y - 7 = 0 \), we can identify:
- \( a_1 = 2 \)
- \( b_1 = 3 \)
- \( c_1 = -7 \)

From the second equation \( (k + 2)x + (3k - 3)y - (5k + 1) = 0 \), we can identify:
- \( a_2 = k + 2 \)
- \( b_2 = 3k - 3 \)
- \( c_2 = -(5k + 1) \)

### Step 3: Set up the condition for infinitely many solutions
For the two equations to have infinitely many solutions, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

### Step 4: Set up the ratios
1. From \( \frac{a_1}{a_2} = \frac{2}{k + 2} \)
2. From \( \frac{b_1}{b_2} = \frac{3}{3k - 3} \)
3. From \( \frac{c_1}{c_2} = \frac{-7}{-(5k + 1)} = \frac{7}{5k + 1} \)

### Step 5: Equate the first two ratios
Setting the first two ratios equal gives:
\[ \frac{2}{k + 2} = \frac{3}{3k - 3} \]

Cross-multiplying:
\[ 2(3k - 3) = 3(k + 2) \]
Expanding both sides:
\[ 6k - 6 = 3k + 6 \]
Rearranging gives:
\[ 6k - 3k = 6 + 6 \]
\[ 3k = 12 \implies k = 4 \]

### Step 6: Verify with the third ratio
Now, we need to check if this value of \( k \) satisfies the third ratio:
\[ \frac{2}{k + 2} = \frac{7}{5k + 1} \]
Substituting \( k = 4 \):
\[ \frac{2}{4 + 2} = \frac{7}{5(4) + 1} \]
This simplifies to:
\[ \frac{2}{6} = \frac{7}{21} \]
Both sides simplify to \( \frac{1}{3} \), confirming that the value \( k = 4 \) satisfies the condition.

### Conclusion
Thus, the value of \( k \) for which the given pair of equations has infinitely many solutions is:
\[ \boxed{4} \]