Question

For what value of k, the pair of linear equations 3x+y=3 and 6x+ky=8 does

not have a solution.​

Answer 1

Step-by-step explanation:

3x+y-3=0 .... eq i-

6x+ky-8=0 .... eq ii-

a1=3 , b1=1 , c1=(-3)

a2=6 , b2=k , c2=(-8)

If a1/a2 = b1/b2 ≠ c1/c2 has no solutions

3/6 = 1/k ≠ (-3)/(-8)

3/6 = 1/k

3k = 6 ....( cross multiplication )

k = 6/3

k = 2

Hope it's helpful...

Thanks.

Answer 2

Given:

the pair of linear equations

3x+y=3 and 6x+ky=8

Find:

Value of k = ?

Solution:

According to question we have

• a1 = 3, b1 = 1, c1 = 3
• a2 = 6, b2 = k, c2 = 8

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

• $\frac{3}{6} =\frac{1}{k} \neq\frac{3}{8}$
• so we get
• $\frac{3}{6}= \frac{1}{k} \\\\k=\frac{6\times 1}{3}$
• k = 2

Hence the value of k is 2