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

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:
a1a2=b1b2=c1c2
2a+b+1=3a+2b+2=-7-4a+b+1
2a+b+1=3a+2b+2=74a+b+1
2a+b+1=3a+2b+2and3a+2b+2=74a+b+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
ab − 1 = 0 and 12a + 12b + 3 = 7a + 14b + 14
ab = 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
∴‚Äč a = 3 and b = 2

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