# If the system of equations has infinitely many solutions, then

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 $b$.