Question:
If the system of equations has infinitely many solutions, then
$2 x+3 y=7$
$2 a x+(a+b) y=28$
(a) $a=2 b$
(b) $b=2 a$
(c) $a+2 b=0$
(d) $2 a+b=0$
Solution:
Given the system of equations are
$2 x+3 y=7$
$2 a x+(a+b) y=28$
For the equations to have infinite number of solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
$a_{1}=2, a_{2}=2 a, b_{1}=3, b_{2}=a+b$
$\frac{2}{2 a}=\frac{3}{a+b}=\frac{7}{28}$
By cross multiplication we have
$\frac{2}{2 a}=\frac{3}{a+b}$
$2(a+b)=2 a(3)$
$2 a+2 b=6 a$
$2 b=6 a-2 a$
$2 b=4 a$
Divide both sides by 2 . we get $b=2 a$
Hence, the correct choice is $b$.