# Find the value(s) of $p$ for the following pair of equations:$3 x-y-5=0$ and $6 x-2 y-p=0$,if the lines represented by these equations are parallel.

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To do:

We have to find the value(s) of $p$ and $q$ for the given pair of equations.

Solution:

(i) 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 $10$.

(ii) 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.

(iii) 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,

$a_1=-3, b_1=5$ and $c_1=-7$

$a_2=2p, b_2=-3$ and $c_2=-1$

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,

$\frac{a_1}{a_2}=\frac{-3}{2p}$

$\frac{b_1}{b_2}=\frac{5}{-3}=-\frac{5}{3}$

Therefore,

$\frac{a_1}{a_2}≠\frac{b_1}{b_2}$

$\frac{-3}{2p}≠\frac{-5}{3}$

$-3(3)≠-5\times2p$

$-9≠-10p$

$p≠\frac{9}{10}$

Therefore, the values of $p$ are all real values except $\frac{9}{10}$.

(iv) 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 $-4$.

(v) The given system of equations can be written as:

$2x + 3y -7=0$

$2px +(p+q)y -28=0$

The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.

The condition for which the above system of equations has infinitely many solutions is

$\frac{a_{1}}{a_{2}} =\frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Comparing the given system of equations with the standard form of equations, we have,

$a_1=2, b_1=3, c_1=-7$ and $a_2=2p, b_2=p+q, c_2=-28$

Therefore,

$\frac{2}{2p}=\frac{3}{p+q}=\frac{-7}{-28}$

$\frac{1}{p}=\frac{1}{4}$

$p=4$

$\frac{3}{p+q}=\frac{1}{4}$

$4\times 3=1(p+q)$

$p+q=12$

$4+q=12$

$q=12-4=8$

The values of $p$ and $q$ for which the given system of equations has infinitely many solutions are $4$ and $8$ respectively.

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