Question

Find the value 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 1

Given, pair of equations

$2x + 3y = 7, \left (a + b + 1 \right ) x

+ \left ( a + 2b + 2 \right ) y = 4 \left (a + b \right ) + 1$

On comparing the given equation with standard form

i.e. $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$,

we get $a_1=2,b_1=3$ and $c_1=-7$

and $a_2 = \left (a + b + 1 \right ), b_2 = 2 \left (a

+ 2b + 2 \right )$and$c_2 = - \left \{ 4 \left (a + b \right ) + 1 \right \} $

For infinitely many solutions,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\frac{a_1}{a_2}=\frac{2}{a+b+1},\frac{b_1}{b_2}=\frac{3}{a+2b+2}$ and $\frac{c_1}{c_2}=\frac{-7}{-\left(4\right(a+b)+1)}$

Here,

$\frac{2}{a+b+1}=\frac{3}{a+2b+2}=\frac{7}{\left(4\right(a+b)+1)}$

$\therefore$ I II III

On taking I and II terms, we get

$\frac{2}{a+b+1}=\frac{3}{a+2b+2}$

$\Rightarrow 2 \left (a + 2b + 2 \right ) = 3\left (

a + b + 1 \right )$

$\Rightarrow 2a+4b+4=3a+3b+3$

$\Rightarrow 2a-3a-3b+4b=3-4$

$\Rightarrow -a+b=-1$

$\Rightarrow a-b=1$......... (1)

On taking I and III terms, we get

$\Rightarrow \frac{2}{a+b+1}=\frac{7}{\{4(a+b)+1\}}$

$\Rightarrow 2\{4(a+b)+1\}=7(a+b+1)$

$\Rightarrow 2(4a+4b+1)=7a+7b+7$

$\Rightarrow 8a-7a+8b-7b=-2+7$

$\Rightarrow a+b=5$......... (2)

Solving $e{q}^n$ (1) and (2), we get

$a-b=1$

$a+b=5$

___________

$2a=6$

$\Rightarrow a=3$

Now, substituting the value of a in $e{q}^n$ (1), we get

$\Rightarrow a-b=1$

$\Rightarrow 3-b=1$

$\Rightarrow b=2$