Question

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

(ii) For which value of k will the following pair of linear equations have no solution?
$3x+y=1$ ; $(2k-1)x+(k-1)y=2k+1$

Answer 1

i) For the pair of linear equations to have infinite number of solutions, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Given equation is $2x+3y=7$

Thus $a_1=2,b_1=3,c_1=-7$

We know $a_2=a-b,b_2=a+b,c_2=-3a-b+2$

Therefore, $\frac{2}{a-b}=\frac{3}{a+b}=\frac{7}{(3a+b-2)}$

$\Rightarrow -6a-2b+4=-7a+7b$

$\Rightarrow a-9b=-4$ ...(1)

and $-9a+3b+6=-7a-7b$

$\Rightarrow -2a+4b=-6$ ...(2)

Solving (1) and (2), we get

$a=5,b=1$

ii) For the pair of linear equations to have infinite number of solutions,

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

Given equation is $3x+y=1$

Thus $a_1=3,b_1=1,c_1=1$

We know $a_2=2k-1,b_2=k-1,c_2=2k+1$

Therefore, $\frac{3}{2k-1}=\frac{1}{(k-1)}\ne \frac{1}{(2k+1)}$

$\Rightarrow 3k-3=2k-1,2k+1\ne k-1\Rightarrow k=2,k\ne -2$

$\therefore whenk=2$ $\text{then equations will have no solution}$