Find the values of a and b for which the following system of linear equations has infinitely many solutions
` 2x + 3y = 7`,
` (a+b + 1 )x + (a + 2b + 2) y = 4 (a + b ) + 1 `

Answer

To find the values of \( a \) and \( b \) for which the system of linear equations has infinitely many solutions, we need to analyze the given equations:

1. \( 2x + 3y = 7 \)
2. \( (a+b+1)x + (a+2b+2)y = 4(a+b) + 1 \)

### Step 1: Rewrite the equations in standard form
The standard form of a linear equation is \( Ax + By + C = 0 \). We will rewrite both equations in this form.

For the first equation:
\[ 2x + 3y - 7 = 0 \]
Here, \( A_1 = 2 \), \( B_1 = 3 \), and \( C_1 = -7 \).

For the second equation:
\[ (a+b+1)x + (a+2b+2)y - (4(a+b) + 1) = 0 \]
This gives us \( A_2 = a+b+1 \), \( B_2 = a+2b+2 \), and \( C_2 = -(4(a+b) + 1) \).

### Step 2: Set up the condition for infinitely many solutions
For the system of equations to have infinitely many solutions, the ratios of the coefficients must be equal:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]

### Step 3: Write the ratios
From our equations, we have:
\[ \frac{2}{a+b+1} = \frac{3}{a+2b+2} = \frac{-7}{-(4(a+b) + 1)} \]

### Step 4: Equate the first two ratios
Starting with the first two ratios:
\[ \frac{2}{a+b+1} = \frac{3}{a+2b+2} \]
Cross-multiplying gives:
\[ 2(a+2b+2) = 3(a+b+1) \]
Expanding both sides:
\[ 2a + 4b + 4 = 3a + 3b + 3 \]
Rearranging terms:
\[ 2a + 4b + 4 - 3a - 3b - 3 = 0 \]
This simplifies to:
\[ -a + b + 1 = 0 \quad \text{(Equation 1)} \]

### Step 5: Equate the first and third ratios
Now, equate the first and third ratios:
\[ \frac{2}{a+b+1} = \frac{7}{4(a+b)+1} \]
Cross-multiplying gives:
\[ 2(4(a+b)+1) = 7(a+b+1) \]
Expanding both sides:
\[ 8(a+b) + 2 = 7(a+b) + 7 \]
Rearranging terms:
\[ 8(a+b) - 7(a+b) + 2 - 7 = 0 \]
This simplifies to:
\[ a + b - 5 = 0 \quad \text{(Equation 2)} \]

### Step 6: Solve the system of equations
Now we have two equations:
1. \( -a + b + 1 = 0 \)
2. \( a + b - 5 = 0 \)

From Equation 1, we can express \( b \) in terms of \( a \):
\[ b = a - 1 \]

Substituting \( b \) into Equation 2:
\[ a + (a - 1) - 5 = 0 \]
This simplifies to:
\[ 2a - 6 = 0 \]
Thus,
\[ a = 3 \]

Now substituting \( a = 3 \) back into the expression for \( b \):
\[ b = 3 - 1 = 2 \]

### Final Answer
The values of \( a \) and \( b \) are:
\[ \boxed{3} \quad \text{and} \quad \boxed{2} \]