Question

Find the values of a and b for which the following system of linear equations has an infinite number of solutions:
$$\begin{gathered} 2x+3y=7 \\ 2ax+\left(a+b\right)y=28 \end{gathered}$$ [CBSE 2001]

Answer 1

The given system of equations can be written as
$$\begin{gathered} 2x+3y-7=0.....\left(i\right) \\ 2ax+\left(a+b\right)y-28=0.....\left(ii\right) \end{gathered}$$
This system is of the form
$$\begin{gathered} a_{1}x+b_{1}y+c_1=0 \\ {a}_{2}x+b_{2}y+c_2=0 \end{gathered}$$
where $a_1=2,b_1=3,c_1=-7\mathrm{and}{a}_2=2a,b_2=a+b,c_2=-28$
For the given system of linear equations to have an infinite number of solutions, we must have
$$\begin{gathered} \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \\ \Rightarrow \frac{2}{2a}=\frac{3}{a+b}=\frac{-7}{-28} \\ \Rightarrow \frac{2}{2a}=\frac{-7}{-28}=\frac{1}{4}\mathrm{and}\frac{3}{a+b}=\frac{-7}{-28}=\frac{1}{4} \\ \Rightarrow a=4\mathrm{and}a+b=12 \end{gathered}$$
Substituting a = 4 in a + b = 12, we get
$4+b=12\Rightarrow b=12-4=8$
Hence, a = 4 and b = 8.