# Find the value(s) of $p$ for the following pair of equations:$-x+p y=1$ and $p x-y=1$,if the pair of equations has no solution.

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Given:

Given pair of linear equations is:

$-x+p y=1$ and $p x-y=1$.

To do:

We have to find the value of $p$ if the given system of equations has no solution.

Solution:

Comparing the given pair of linear equations with the standard form of linear equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$, we get,

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

$a_2=p, b_2=-1$ and $c_2=-1$

A system of equations has no solution if the lines are parallel to each other.

Here,

$\frac{a_1}{a_2}=\frac{-1}{p}$

$\frac{b_1}{b_2}=\frac{p}{-1}$

$\frac{c_1}{c_2}=\frac{-1}{-1}=1$

Therefore,

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

$\frac{-1}{p}=-p$

$-p^2=-1$

$p^2=1$

$p=\sqrt1$

$p=\pm 1$

$\frac{a_1}{a_2}≠\frac{c_1}{c_2}$

$\frac{-1}{p}≠1$

$p≠-1$

This implies,

$p=1$

Therefore, the value of $p$ is 1.

Updated on 10-Oct-2022 13:27:14