Question

Find the value of p and q for which following system of linear equation have infinite number of solution 2x-3y=7;(p+q)x-(p+q-3)y=4p+q

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of linear equation is

2x - 3y = 7

and

(p + q)x - (p + q - 3)y = 4p + q

Further it is given that,

System of Linear Equations have infinitely many solutions.

We know,

The pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 have

$\boxed{ \rm \: infinite \: solutions \: when \: \rm \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}}$

So, on comparing the given two equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we get  

• a₁ = 2

• b₁ = - 3

• c₁ = 7

• a₂ = p + q

• b₂ = - (p + q - 3)

• c₂ = 4p + q

So, on substituting the values in

$\rm :\longmapsto\: \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

$\rm :\longmapsto\:\dfrac{2}{p + q} = \dfrac{ - 3}{ - (p + q - 3)} = \dfrac{7}{4p + q}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{2}{p + q} = \dfrac{ 3}{(p + q - 3)} = \dfrac{7}{4p + q}$

Taking first and second member, we have

$\rm :\longmapsto\:\dfrac{2}{p + q} = \dfrac{ 3}{(p + q - 3)}$

$\rm :\longmapsto\:2p + 2q - 6 = 3p + 3q$

$\rm :\longmapsto\:p + q = - \: 6 - - - (3)$

Taking first and third member,

$\rm :\longmapsto\:\dfrac{2}{p + q} = \dfrac{7}{4p + q}$

$\rm :\longmapsto\:8p + 2q = 7p + 7q$

$\rm :\longmapsto\:8p - 7p =7q - 2q$

$\rm :\longmapsto\:p = 5q - - - (4)$

Substituting the value of p in equation (3), we get

$\rm :\longmapsto\:5q + q = - 6$

$\rm :\longmapsto\:6q= - 6$

$\bf :\longmapsto\:q= - 1$

On substituting the value of q in equation (4), we get

$\rm :\longmapsto\:p = 5( - 1)$

$\bf :\longmapsto\:p = - 5$

$\red{\begin{gathered}\begin{gathered}\bf\: Hence \: -\begin{cases} &\sf{p \: = \: - \: 5} \\ &\sf{q \: = \: - \: 1} \end{cases}\end{gathered}\end{gathered}}$

Additional Information :-

The pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 is

1. Consistent having unique solution iff

$\rm :\longmapsto\: \dfrac{a_1}{a_2} \: \ne \: \dfrac{b_1}{b_2}$

2. Inconsistent having no solution iff

$\rm :\longmapsto\: \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \: \ne \: \dfrac{c_1}{c_2}$