Question

Find the values of a and b for which the following system of linear equations has an infinite number of solutions:
$$\begin{gathered} 2x+3y=7 \\ \left(a+b\right)x+\left(2a-b\right)y=21 \end{gathered}$$ [CBSE 2001]

Answer 1

The given system of equations can be written as
$$\begin{gathered} 2x+3y-7=0.....\left(i\right) \\ \left(a+b\right)x+\left(2a-b\right)y-21=0.....\left(ii\right) \end{gathered}$$
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 $a_1=2,b_1=3,c_1=-7\mathrm{and}{a}_2=a+b,b_2=2a-b,c_2=-21$
For the given system of linear equations to have an infinite number of solutions, we must have
$$\begin{gathered} \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \\ \Rightarrow \frac{2}{a+b}=\frac{3}{2a-b}=\frac{-7}{-21} \\ \Rightarrow \frac{2}{a+b}=\frac{-7}{-21}=\frac{1}{3}\mathrm{and}\frac{3}{2a-b}=\frac{-7}{-21}=\frac{1}{3} \\ \Rightarrow a+b=6\mathrm{and}2a-b=9 \end{gathered}$$
Adding $a+b=6\mathrm{and}2a-b=9$, we get
$3a=15\Rightarrow a=\frac{15}{3}=5$
Now, substituting a = 5 in a + b = 6, we have
$5+b=6\Rightarrow b=6-5=1$
Hence, a = 5 and b = 1.