Question

For what value of 'p' the pair of linear equations $\left(p+2\right)x-\left(2p+1\right)y=3\left(2p-1\right)$ and $2x-3y=7$ will be have a unique solution.

Answer 1

The given pair of linear equation :

$\Rightarrow$ $(p+2)x-(2p+1)y=3(2p-1)$ ------ ( 1 )

$\Rightarrow$ $2x-3y=7$ ------- ( 2 )

On comparing with equations,

$a_{1}x+b_{1}y+c_1=0$

and $a_{2}x+b_{2}y+c_2=0$

We get, $a_1=(p+2),b_1=-(2p+1),c=-\left[3\right(2p-1\left)\right]$

$a_2=2,b_2=-3,c=-7$

$\Rightarrow$ $\frac{a_1}{a_2}=\frac{p+2}{2}$ and $\frac{b_1}{b_2}=\frac{-(2p+1)}{-3}$

A pair of linear equations has a unique solution, if

$\Rightarrow$ $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

$\Rightarrow$ $\frac{(p+2)}{2}\ne \frac{-(2p+1}{-3}$

$\Rightarrow$ $-3(p+2)\ne 2(-2p-1)$

$\Rightarrow$ $-3p-6\ne -4p-2$

$\Rightarrow$ $-3p+4p\ne -2+6$

$\Rightarrow$ $p\ne 4$

Therefore, the given system will have unique solution for all real values of $p$ other than $4$.