Question

The value of ‘k’ so that the system of equations 3x – 4y – 7 = 0 and 6x – ky – 5 = 0 have a unique solution is


k≠−8
k≠8
k≠4
k≠−4

Answer 1

Answer:

Given system of equations are

6x - 2y = 3

6x - 2y - 3 = 0 ----( 1 )

kx - y = 2

kx - y - 2 = 0 ----( 2 )

Compare above equations with

a1 x + b1 y + c1 = 0 and

a2 x + b2 y + c2 = 0 , we get

a1 = 6 , b1 = -2 , c1 = -3 ;

a2 = k , b2 = -1 , c2 = -2 ;

Now ,

a1/a2 ≠ b1/b2

[ Given they have Unique solution ]

6/k ≠ ( -2 )/( -1 )

6/k ≠ 2

k/6 ≠ 1/2

k ≠ 6/2

k ≠ 3

Therefore ,

For all real values of k , except k≠ 3,

Above equations has unique solution.

Answer 2

AnsWer :

k≠8.

Concepts Required :

• Many solution.
• No solution.
• Unique Solution.

Solution :

We have, Equation

3x – 4y – 7 = 0 ________( 1 )

6x – ky – 5 = 0________( 2 )

Let Compare With, Unique Solution ( Condition )

$\tt \mapsto \dfrac{ a_1}{ a_2} \neq \dfrac{b_1}{b_2 }$

Where as,

• a1 = 3.
• a2 = 6.
• b1 = -4.
• b2= -k.

Putting, given value in it.

$\tt \mapsto \dfrac{3}{6} \neq \dfrac{ - 4}{ - k}$

$\tt\mapsto \dfrac{1}{2} \neq \dfrac{4}{ k}$

$\tt \mapsto \red{k \neq 8.}$

Therefore, the value of k not equal to 8.

$\rule{200}3$

Some information :

Condition,Many solution.

• a1 /a2 = b1 /b2 = c1/c2

Condition,No solution.

• a1 /a2 = b1 /b2 ≠ c1/c2

Condition,Unique Solution.

• a1 /a2 ≠ b1 /b2.