Question

If the system of equations
$2x+3y=7$
$2ax+(a+b)y=28$
has infinitely many solutions, then ______.

• A. $a=2b$
• B. $b=2a$
• C. $a+2b=0$
• D. $2a+b=0$

Answer 1

Answer: (B)

$b=2a$

Given : $2x+3y=7$

$2ax+(a+b)y=28$

Comparing $2x+3y=7$ with $a_{1}x+b_{1}y+c_1=0$, we get,

$a_1=2,b_1=3,c_1=-7$

Comparing $2ax+(a+b)y=28$ with $a_{2}x+b_{2}y+c_2=0$

$a_2=2a,b_2=(a+b),c_2=-28$

For infinitely many solutions, we know,

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

$\frac{2}{2a}=\frac{3}{a+b}=\frac{7}{28}$

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

$⟹6a=2a+2b$

$⟹4a=2b$

$⟹2a=b$