Question

For which values of $a$ and $b$, will the following pair of linear equations have infinitely many solutions? $x+2y=1$ and $(a-b)x+(a+b)y=a+b-2$

Answer 1

Step 1: Convert the equations to standard form

Given two lines are,

$x+2y=1$ and $(a-b)x+(a+b)y=a+b-2$

The standard linear equation is given as,

$ax+by+c=0$

Then, the two equations in standard form are given as,

$x+2y=1\Rightarrow x+2y-1=0$

$(a-b)x+(a+b)y=a+b-2\Rightarrow (a-b)x+(a+b)y-a-b+2=0$

Step 2: Compare the equations with the standard equation

Comparing the two equations with the standard equation we define,

$a_1=1$, $b_1=2$ and $c_1=-1$

$a_2=a-b$, $b_2=a+b$ and $c_2=-(a+b-2)$

Step 3: Defining the condition for infinite solutions

For infinitely many solutions of the pair of linear equations condition is satisfied by,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Thus,

$\frac{1}{a-b}=\frac{2}{a+b}=\frac{1}{a+b-2}$

We have,

$\begin{array}{rcl}& & \begin{array}{cc}& \frac{1}{a-b}=\frac{2}{a+b}\\ \Rightarrow & 2(a-b)=a+b\\ \Rightarrow & a=2b\end{array}\end{array}$

We also have,

$\begin{array}{cc}& \frac{2}{a+b}=\frac{1}{a+b-2}\\ \Rightarrow & a+b=2(a+b-2)\\ \Rightarrow & a+b=4\end{array}$

Substituting the above equation in the previous equation,

$$\begin{gathered} 2b+b=4 \\ \Rightarrow b=1 \end{gathered}$$

Thus, $a=2\times 1=2$

Therefore, when $a=2$ and $b=1$, the given set of linear equations has infinitely many solutions.