Question

For the pair of equations θx+3y + 7 =0 and 2x + 6y -14 =0, to have infinitely many solutions, the value of θ should be 1. Is the statement true? Give reasons.​

Answer 1

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

Given pair of lines are

$\rm :\longmapsto\:\theta x + 3y + 7 = 0 - - - (1)$

and

$\rm :\longmapsto\:2x + 6y - 14 = 0 - - - (2)$

Now,

$\red{\rm :\longmapsto\:When \: \theta \: = \: 1}$

The equation (1) can be rewritten as

$\rm :\longmapsto\: x + 3y + 7 = 0$

So, pair of lines we have

$\rm :\longmapsto\: x + 3y + 7 = 0$

and

$\rm :\longmapsto\: 2x + 6y - 14= 0$

We know

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

$\boxed{ \tt{ \: 1) \: no \: solution \: when \:   \rm \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \: }}$

$\boxed{ \tt{ \: 2) infinite \: solutions \: when \: \rm \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \: }}$

$\boxed{ \tt{ \: 3) unique \: solution \: when \: \rm \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \: }}$

Now,

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

• a₁ = 1

• b₁ = 3

• c₁ = 7

• a₂ = 2

• b₂ = 6

• c₂ = - 14

So,

$\red{\rm :\longmapsto\:\dfrac{a_1}{a_2} = \dfrac{1}{2} \: }$

$\red{\rm :\longmapsto\:\dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2} \: }$

$\red{\rm :\longmapsto\:\dfrac{c_1}{c_2} = - \dfrac{7}{14} = - \dfrac{1}{2} \: }$

$\red{\rm \rm \implies\: \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \: }$

It means, Pair of equations have no solution.

• So, given statement is False.