Question

For which values of $a$ and $b$ does the following pair of linear equations have an infinite number of solutions? $2x+3y=7,\left(a-b\right)x+\left(a+b\right)y=3a+b-2$.

Answer 1

Step 1: Calculate the ratios of values of coefficients of the linear equations.

Given equations are:

$\begin{array}{rcl}2x+3y& =& 7\\ \left(a-b\right)x+\left(a+b\right)y-\left(3a+b-2\right)& =& 0\end{array}$

On solving both the equations,

$$\begin{gathered} \frac{a_1}{a_2}=\frac{2}{\left(a-b\right)} \\ \frac{b_1}{b_2}=\frac{3}{\left(a+b\right)} \\ \frac{c_1}{c_2}=\frac{-7}{-\left(3a+b-2\right)} \end{gathered}$$

Step 2: Applying the conditions for infinitely many solutions.

The condition for infinitely many solutions is $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

So, $\frac{2}{\left(a-b\right)}=\frac{3}{\left(a+b\right)}$

On simplifying the above equation,

$$\begin{gathered} \Rightarrow 2\left(a+b\right)=3\left(a-b\right) \\ \Rightarrow 2a+2b=3a-3b \\ \Rightarrow a=5b \end{gathered}$$

and $\frac{2}{\left(a-b\right)}=\frac{-7}{-\left(3a+b-2\right)}$

On simplifying the above equation,

$$\begin{gathered} \Rightarrow -2\left(3a+b-2\right)=-7\left(a-b\right) \\ \Rightarrow -6a-2b+4=-7a+7b \\ \Rightarrow a=9b-4 \end{gathered}$$

Step 3: Substituting $a=5b$in the above equation

$$\begin{gathered} \Rightarrow 5b=9b-4 \\ \Rightarrow 4b=4 \\ \Rightarrow b=1 \end{gathered}$$

and

$\begin{array}{rcl}& \Rightarrow & a=5\left(1\right)\\ & =& 5\end{array}$

Hence, values of $a$ and $b$are $5\&1$.