# Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:$2x+3y=7,\phantom{\rule{0ex}{0ex}}\left(a+b+1\right)x+\left(a+2b+2\right)y=4\left(a+b\right)+1.$

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:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
$\frac{2}{\left(a+b+1\right)}=\frac{3}{\left(a+2b+2\right)}=\frac{-7}{-\left[4\left(a+b\right)+1\right]}$
$⇒\frac{2}{\left(a+b+1\right)}=\frac{3}{\left(a+2b+2\right)}=\frac{7}{4\left(a+b\right)+1}$
$⇒\frac{2}{\left(a+b+1\right)}=\frac{3}{\left(a+2b+2\right)}\mathrm{and}\frac{3}{\left(\mathrm{a}+2\mathrm{b}+2\right)}=\frac{7}{4\left(a+b\right)+1}$

⇒ 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
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
∴​ a = 3 and b = 2

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