Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
$$\begin{gathered} 2x+3y=7, \\ (a+b+1)x+(a+2b+2)y=4(a+b)+1. \end{gathered}$$

The given system of equations:
2x + 3y = 7
⇒ 2x + 3y − 7 = 0                                   ....(i)
And, (a + b + 1)x + (a + 2b + 2)y = 4(a + b) + 1
⇒ (a + b + 1)x + (a + 2b + 2)y − [4(a + b) + 1] = 0    ....(ii)          
These equations are of the following form:
a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 2, b₁=  3, c₁ = −7 and a₂ = (a + b + 1), b₂ = (a + 2b + 2), c₂ = −[4(a + b) + 1]
For an infinite number of solutions, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{2}{\left(a+b+1\right)}=\frac{3}{\left(a+2b+2\right)}=\frac{-7}{-\left[4\left(a+b\right)+1\right]}$
$\Rightarrow \frac{2}{\left(a+b+1\right)}=\frac{3}{\left(a+2b+2\right)}=\frac{7}{4\left(a+b\right)+1}$
$\Rightarrow \frac{2}{\left(a+b+1\right)}=\frac{3}{\left(a+2b+2\right)}\mathrm{and}\frac{3}{\left(\mathrm{a}+2\mathrm{b}+2\right)}=\frac{7}{4\left(a+b\right)+1}$

⇒ 2(a + 2b + 2) = 3(a + b+ 1) and 3[4(a + b) + 1] = 7(a + 2b + 2)
⇒ 2a + 4b + 4 = 3a + 3b + 3 and 3(4a + 4b + 1) = 7a + 14b + 14
⇒ a − b − 1 = 0 and 12a + 12b + 3 = 7a + 14b + 14
⇒ a − b = 1 and 5a − 2b = 11
a = (b + 1)     ....(iii)
5a − 2b = 11     ....(iv)
On substituting a = (b + 1) in (iv), we get:
5(b + 1) − 2b = 11
⇒ 5b + 5 − 2b = 11
⇒ 3b = 6
⇒ b = 2
On substituting b = 2 in (iii), we get:
a = 3
∴​ a = 3 and b = 2