Question

the value of the k for which the pair of equation kx-y=2 And 6x-2y=3 will have many infinately solutions

Answer 1

GIVEN:

• Kx – y = 2
• 6x – 2y = 3

TO FIND:

• What is the value of k ?

SOLUTION:

We have given two Equations

• Kx–y = 2
• 6x–2y = 3

We know that, the condition for infinitely many solution is:-

$\large{\boxed{\bf{\star \: \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \: \star}}}$

Where,

• $\sf{ a_1 = k}$
• $\sf{ a_2 = 6}$
• $\sf{ b_1 = -1}$
• $\sf{ b_2 = -2}$
• $\sf{ c_1 = -2}$
• $\sf{ c_2 = -3}$

According to question:-

On putting the given values in the formula, we get

$\sf{\longmapsto \dfrac{k}{6} = \dfrac{-1}{-2} = \dfrac{-2}{-3}}$

$\sf{\longmapsto \dfrac{k}{6} = \dfrac{-1}{-2} }$

Use cross product

$\sf{\longmapsto k \times -2 = 6 \times -1 }$

$\sf{\longmapsto -2k = -6 }$

$\sf{\longmapsto k = \cancel\dfrac{-6}{-2} }$

$\bf{\longmapsto k = 3}$

❝ Hence, the value of k is 3 ❞

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Answer 2

$\huge\bold\red{Answer:-}$

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Given:-

• linear equation kx-y=2 and 6x-2y=3
• equations have infinitely many solution

To find:-

• value of k

Solution:-

Given equations are

kx-y=2 and 6x-2y=3

We know that ,

If two equations have infinitely many solution,then

$\frac{a1}{a2}$=$\frac{b1}{b2}$=$\frac{c1}{c2}$

Here,,

a1=k

a2=6

b1=(-1)

b2=(-2)

c1=2

c2=3

putting the value we get,

$\frac{k}{6}$=$\frac{(-1)}{(-2)}$=$\frac{2}{3}$

$\frac{k}{6}$=$\frac{(-1)}{(-2)}$

=> (-2)k=(-6)

=>k=(-6)/(-2)

=>k=$\cancel\dfrac{-6}{-2}$

=>k=3

•°• required value of k is 3

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$\huge\bold\green{Thanks}$