Question

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

• A. $a=3$ and $b=2$
• B. $a=2$ and $b=1$
• C. $a=0$ and $b=5$
• D. $a=6$ and $b=0$

Answer 1

The correct option is A $a=3$ and $b=2$

We have $2x+3y=7\Rightarrow 2x+3y-7=0$ and $(a+b+1)x+(a+2b+2)y::::4(a+b)+1$
$\Rightarrow (a+b+1)x+(a+2b+2)y-\left\{4\right(a+b)+1\}=0$
The required condition for an infinite number of
solution is $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\therefore \frac{2}{a+b+1}=\frac{3}{a+2b+2}=\frac{-7}{-\left\{4\right(a+b)+1\}}$
$\Rightarrow \frac{1}{a+b+1}=\frac{3}{a+2b+2}$
$\Rightarrow 2a+4b+4=3a+3b+3$
and $\frac{3}{a+2b+2}=\frac{7}{4(a+b)+1}$
and $12a+12b+3=7a+14b+14$
$\Rightarrow a-b-1=0$ and $5a-2b=11$
$\Rightarrow a-b=1$...(1)
and $5a-2b=11$ ...(2)
Multiplying (1) by 2 we get $2a-2b=2$ ...(3)
Subtracting (3) from (2) we get $3a=9$
$\Rightarrow a=\frac{9}{3}=3$
Put $a=3$ in (1), we get $3-b=1\Rightarrow b=2$
Hence, the given system of equations will have infinite number of solutions when $a=3$ and $b=2$.