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

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

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

Solution:

We can write the given system of equations as
$2 x+3 y=7$
$\Rightarrow 2 x+3 y-7=0\dots \dots(i)$
and $(a+b+1) x-(a+2 b+2) y=4(a+b)+1$
$(a+b+1) x-(a+2 b+2) y-[4(a+b)+1]=0\dots \dots(i)$
The given equations are of the following form:
$a_{1} x+b_{1} y+c_{1}=0, a_{2} x+b_{2} y+c_{2}=0$
where, $a_{1}=2, b_{1}=3, c_{1}=-7$ and $a_{2}=(a+b+1), b_{2}=(a+2 b+2), c_{2}=-[4(a+b)+1]$

For an infinite no. of solutions, we must have:
$\begin{array}{l}\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \\ \frac{2}{(a+b+1)}=\frac{3}{(a+2 b+2)}=\frac{-7}{-[4(a+b)+1]} \\ \Rightarrow \frac{2}{(a+b+1)}=\frac{3}{(a+2 b+2)}=\frac{7}{[4(a+b)+1]} \\ \Rightarrow \frac{2}{(a+b+1)}=\frac{3}{(a+2 b+2)} \text { and } \frac{3}{(a+2 b+2)}=\frac{7}{[4(a+b)+1]} \\ \Rightarrow 2(a+2 b+2)=3(a+b+1) \text { and } 3[4(a+b)+1]=7(a+2 b+2) \\ \Rightarrow 2 a+4 b+4=3 a+3 b+3 \text { and } 3(4 a+4 b+1)=7 a+14 b+14 \\ \Rightarrow a-b-1=0 \text { and } 12 a+12 b+3=7 a+14 b+14 \\ \Rightarrow a-b=1 \text { and } 5 a-2 b=11 \\ a=(b+1) \quad \ldots \ldots \text { (iii) } \\ 5 a-2 b=11 \ldots \ldots \text { (iv) }\end{array}$
On substituting $\mathrm{a}=(\mathrm{b}+1)$ in equation(iv), we obtain:
$\begin{array}{l} 5(b+1)-2 b=11 \\ \Rightarrow 5 b+5-2 b=11 \\ \Rightarrow 3 b=6 \\ \Rightarrow b=2 \end{array}$
On substituting $\mathrm{b}=2$ in equation(iii), we obtain:
$\mathrm{a}=3$
$\therefore \mathrm{a}=3$ and $\mathrm{b}=2$