Find the value of $p$ such that the pair of equations $ 3 x-y-5=0 $ and $ 6 x-2 y-p=0 $, has no solution.

Given:

Given pair of linear equations is:

\( 3 x-y-5=0 \) and \( 6 x-2 y-p=0 \).

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,

$a_1=3, b_1=-1$ and $c_1=-5$

$a_2=6, b_2=-2$ and $c_2=-p$

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

Here,

$\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{-1}{-2}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{-5}{-p}$

Therefore,

$\frac{a_1}{a_2}≠\frac{c_1}{c_2}$

$\frac{1}{2}≠\frac{5}{p}$

$p≠5\times2$

$p≠10$

Therefore, the values of $p$ are all real values except 10.

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Updated on: 10-Oct-2022

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