Question

find the value of k for which the following pair of equations have unique solutions 7x-2y=3,22x-ky=8​

Answer 1

Step-by-step explanation:

condition for unique solutions is a1/a2 ≠ b1/b2

7x-2y=3,22x-ky=8

a1=7,b1=-2 and a2=22,b2=-k

7/22≠-2/-k

7k≠44

k≠44/7

Answer 2

Answer:

All the values of 'k' other than  k = $\frac{44}{7}$ , the given pair linear equations have unique solutions

Step-by-step explanation:

Given,

The pair of linear equations 7x-2y=3 and 22x-ky=8​ have unique solutions

To find,

The value of 'k'

Recall the concept,

A pair of linear equations of the form $a_1x+b_1y+c_1 = 0$ and $a_2x+b_2y+c_2 = 0$  have unique solutions if

$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

Solution:

We have the equations

7x-2y=3 and 22x-ky=8

Comparing with​ $a_1x+b_1y+c_1 = 0$  and  $a_2x+b_2y+c_2 = 0$  we get

$a_1 = 7, b_1 = -2 \\a_2 = 22, b_2 = -k$

$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$  ⇒$\frac{7}{22} \neq \frac{-2}{-k}$

7k≠  44

$k \neq \frac{44}{7}$

All the values of 'k' other than k = $\frac{44}{7}$ , the given pair linear equations have unique solutions

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