Question

If the system of equations $2x+3y=7$ and $(a+b)x+(2a-b)y=21$ has infinitely many solutions, then

• A. $a=1,b=5$
• B. $a=5,b=1$
• C. $a=-1,b=+5$
• D. $a=5,b=-1$

Answer 1

Answer: (B)

$a=5,b=1$

$2x+3y=7$

$(a+b)x+(2a-b)y=21$

Condition for infinitely many solutions is
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\Rightarrow \frac{2}{a+b}=\frac{3}{2a-b}=\frac{7}{21}$
when $\frac{2}{a+b}=\frac{1}{3}\Rightarrow a+b=6$ .........(i)
when $\frac{3}{2a-b}=\frac{1}{3}\Rightarrow 2a-b=9$ ...........(ii)

Adding eq (i) and (ii)

$3a=15$
we get $a=5$ and

$5+b=6$

$b=1.$
Hence the correct option is (B).