Question

For which value of k will the pair of linear equations 3x+y=1 and 3xk+5y=2

have no solutions?​

Answer 1

SOLUTION

TO DETERMINE

The value of k will the pair of linear equations 3x + y = 1 and 3xk + 5y = 2 have no solutions

CONCEPT TO BE IMPLEMENTED

For the given two linear equations

$\displaystyle \sf{ a_1x+b_1y+c_1=0 \: and \: \: a_2x+b_2y+c_2=0}$

Consistent :

One of the Below two condition is satisfied

1. Unique solution :

$\displaystyle \sf{ \: \frac{a_1}{a_2} \ne \frac{b_1}{b_2} }$

2. Infinite number of solutions :

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} = \: \frac{c_1}{c_2}}$

Inconsistent :

No solution :

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \: \frac{c_1}{c_2}}$

EVALUATION

Here the given system of equations are

3x + y = 1 - - - - - - - (1)

3xk + 5y = 2 - - - - - - (2)

Comparing with the equation

a₁x + b₁y = c₁ and a₂x + b₂y = c₂ we get

a₁ = 3 , b₁ = 1 , c₁ = 1 and a₂ = 3k , b₂ = 5 , c₂ = 2

Since the pair of linear equations have no solutions

So we have

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \: \frac{c_1}{c_2}}$

$\displaystyle \sf{ \implies \frac{3}{3k} = \frac{1}{5} \ne \: \frac{1}{2}}$

$\displaystyle \sf{ \implies \frac{3}{3k} = \frac{1}{5} }$

$\displaystyle \sf{ \implies \frac{1}{k} = \frac{1}{5} }$

$\displaystyle \sf{ \implies k = 5 }$

FINAL ANSWER

Hence the required value of k = 5

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