Question

for what value of k, the following fair of linear Equation has no Solution ? 3x + ky = 11 , 6x + 2y = 18​

Answer 1

Answer:

The value of k = 10.

Step-by-step explanation:

Given two equations, 3x – y – 5 = 0 (i.e) 3x - y = 5 …. (1)

$\sf \: 6x - 2y – k = 0 (i.e) \: 6x - 2y = k …. (2)$

Divide equation (2) by 2,

$\sf \: it \: becomes, 3x – y = \left(\frac{k}{2}\right)… (3)$

From equation ,(1) and (3) LHS are equal hence consider RHS

$\sf \: (\frac{k}{2}) = 5, k = 10 \\ \sf \: for \: the \: problem \: \: to \: have \: solution,$

$\sf \: Condition \: for \: no \: solution \: be \: \bold{ k ≠ 10.}$

Answer 2

$\sf\huge\bold{Answer:}$

Given :

3x + ky = 11 , 6x + 2y = 18

Equations have no solution

To find :

Value of k

Solution :

When equations have no Solution

$\boxed{\tt \red{ \frac{a_1 }{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2} } }$

Coverting given equations into general form ax+by+c=0

3x + ky = 11

→ 3x+ky-11=0...[ 1 ]

6x + 2y = 18

→6x + 2y - 18=0...[ 2 ]

From [ 1 ] and [ 2 ]

$\tt a_1$ = 3 [coefficient of x]

$\tt b_1$ = k [coefficient of y]

$\tt c_1$ = -11[constant term]

$\tt a_2$ = 6 [coefficient of x]

$\tt b_2$ = 2 [coefficient of y]

$\tt c_2$ = -18[constant term]

• Inserting values

$\tt \: \frac{3}{6} = \frac{k}{2} ≠\frac{ - 11}{ - 18}$

$\tt \: \cancel{ \frac{3}{6} } \: \: = \: \: \frac{k}{2} \: ≠ \: \: \frac{ \cancel{-} 11}{ \cancel{-} 18}$

$\tt \: \frac{1}{2} = \frac{k}{2} ≠\frac{ 11}{ 18}$

$\tt\ \frac{1}{2} = \frac{k}{2}$

$\tt\ 2 \times k = 1 \times 2$

$\tt\ 2k = 2$

$\tt\ k = \frac{2}{2}$

$\tt\ k = \cancel\frac{2}{2}$

$\tt\ k = 1$

$\huge\boxed {{\underline \purple{k = 1} }}$