Question

What will be the value of k if equations 3x-y=4 and kx+5y=3 have unique solution .
(A) K=15
(B)k≠15
(C) k≠-15
(D) None of these​

Answer 1

Answer:

$\bigstar{\bold{Option\:C:k\neq -15}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

Two equations:

• 3x - y = 4
• kx + 5y = 3

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• The value of the k if the pair of the equations has a unique solution

$\Large{\underline{\underline{\bf{Solution:}}}}$

➻ We have to find the value of k.

➻ It is given that the pair of equations has a unique solution.

➻ Hence,

   $\sf{\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} }$

  where a₁ = 3, a₂ = k, b₁ = 3, b₂ = 5

➻ Substituting the data,

   $\sf{\dfrac{3}{k} \neq \dfrac{-1}{5} }$

→ Cross multiplying we get the value for k

   -1 × k ≠ 3 × 5

    -k ≠ 15

     k ≠ -15

➻ Hence k can take the value of any real number except -15.

➻ That is the pair of equations will have a unique solution for all real values of k except -15

   $\boxed{\bold{k\neq -15}}$

➻ Hence option C is correct

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ If a pair of equations

    a₁x + b₁y + c₁ = 0

    a₂x + b₂y + c₂ = 0

➵ has a unique solution and is consistent,

     $\sf{\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} }$

➵ has infinte solutions and is consistent,

     $\sf{\dfrac{a_1}{a_2} =\dfrac{b_1}{b_2} =\dfrac{c_1}{c_2} }$

➵  has no solution and is inconsistent,

     $\sf{\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq \dfrac{c_1}{c_2} }$

Answer 2

Answer :-

★ Option C) K ≠ -15

The required answer will be Option C, when both the given linear equations will have unique solution.

• ★ Concept : Simple the concept used here is of the fractional equivalency or non - equivalency of coefficients of each term of Linear Equations.

★ Solution :

Given, linear equations :-

• 3x - y = 4 ..... (i)

• kx + 5y = 3 .... (ii)

For unique solutions in linear equations, the condition is,

$\: \: \: \: ( \frac{a_{1}}{a_{2}}) \: \not = \: ( \frac{b_{1} }{b_{2}})$

This ⬆ equation also gives interesting lines when drawn in graph.

According to equation, a and b are coefficients of both the equations respectively.

By applying values, we get,

$\frac{3}{k} \: \: \: \not = \: \: \: \frac{ - 1}{5}$

On cross multiplication, of both terms using inequality, we get,

=> - K ≠ 15

=> K ≠ -15

Hence, we get our required answer, that is K ≠ -15.

★ More to Know :-

• On basis of variables used in equations, Linear Equations are classified into, One Variable, Two Variable and so on..

• The one used here is known as Linear Equation in Two Variable. The graph of the solutions of these equations, intersect both x - axis and y - axis.

• In the Linear Equations in Single / One Variable, only one variable term is used and equation will be only one. The graph of the solution will either intersect x - axis or y - axis.