Question

Check whether the following pair of

linear equation has unique solution, no

solution or infinitely many solutions. In

case there is a unique solution find it by

using a cross multiplication method.

x - 3y - 7 = 0

3x - 3y -15 =0
​

Answer 1

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

Given pair of lines are

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

can be rewritten as

$\rm :\longmapsto\:x - 3y = 7$

and

$\rm :\longmapsto\:3x - 3y = 15$

can be rewritten as

$\rm :\longmapsto\:x - y = 5$

So, pair of linear equations in simplest form is

$\rm :\longmapsto\:x - 3y = 7$

and

$\rm :\longmapsto\:x - y = 5$

We know,

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

$\rm :\longmapsto\:1) \: \: unique \: solution \: when \:   \rm \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$

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

$\rm :\longmapsto\:3) \: \: no \: solution \: when \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

So, 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 = 1 , b₂ = - 1 , c₂ = 5

So,

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

and

$\rm :\longmapsto\:\dfrac{b_1}{b_2} = \dfrac{ - 3}{ - 1} = 3$

$\bf\implies \:\dfrac{a_1}{a_2} \: \ne \: \dfrac{b_1}{b_2}$

System of equations have unique solution.

Now,

Solution using Cross Multiplication method, we have

$\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf 2 & \bf 3 & \bf 1& \bf 2\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf - 3 & \sf 7 & \sf 1 & \sf - 3\\ \\ \sf - 1 & \sf 5 & \sf 1 & \sf - 1\\ \end{array}} \\ \end{gathered}$

So,

$\rm :\longmapsto\:\dfrac{x}{ - 15 - ( - 7)} = \dfrac{y}{7 - 5} = \dfrac{ - 1}{ - 1 - ( - 3)}$

$\rm :\longmapsto\:\dfrac{x}{ - 15 + 7} = \dfrac{y}{7 - 5} = \dfrac{ - 1}{ - 1 + 3}$

$\rm :\longmapsto\:\dfrac{x}{ - 8} = \dfrac{y}{2} = \dfrac{ - 1}{2}$

Taking first and third member, we have

$\rm :\longmapsto\:\dfrac{x}{ - 8} = \dfrac{ - 1}{2}$

$\bf\implies \:x = 4$

Taking second and third member, we have

$\rm :\longmapsto\: \dfrac{y}{2} = \dfrac{ - 1}{2}$

$\bf\implies \:y = - 1$