Find the value(s) of $ p $ for the following pair of equations:
$ 2 x+3 y-5=0 $ and $ p x-6 y-8=0 $,
if the pair of equations has a unique solution.

Given:

Given pair of linear equations is:

\( 2 x+3 y-5=0 \) and \( p x-6 y-8=0 \).

To do:

We have to find the value of $p$ if the given system of equations has a unique 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,

$a_1=2, b_1=3$ and $c_1=-5$

$a_2=p, b_2=-6$ and $c_2=-8$

A system of equations has a unique solution if it satisfies the following condition,

$\frac{a_1}{a_2}≠ \frac{b_1}{b_2}$

Here,

$\frac{a_1}{a_2}=\frac{2}{p}$

$\frac{b_1}{b_2}=\frac{3}{-6}=-\frac{1}{2}$

Therefore,

$\frac{a_1}{a_2}≠\frac{b_1}{b_2}$

$\frac{2}{p}≠\frac{-1}{2}$

$2(2)≠-1\times p$

$4≠-p$

$p≠-4$

Therefore, the values of $p$ are all real values except $-4$.

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Updated on: 10-Oct-2022

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