Question

Find the value of k for which the pair of linear equation 8x+5y=9 and kx+10y=15 are inconsistent

Answer 1

Given as :

The pair of linear equation are for inconsistent solution

8 x + 5 y = 9

k x + 10 y = 15

To Find :

The value of k

Solution :

Let the equations are

$a_1x + b_1y+c_1 = 0$

and

$a_2x+b_2y+c_2=0$

For inconsistent solution , i.e for infinite solution

$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

i.e  by comparing

$\dfrac{8}{k}$  =  $\dfrac{5}{10}$  =  $\dfrac{-9}{-15}$

Or,  $\dfrac{8}{k}$  =  $\dfrac{5}{10}$

By cross multiplication

8 × 10 = 5 × k

Or , 5 k = 80

∴        k = $\dfrac{80}{5}$

i.e       k = 16

And

$\dfrac{8}{k}$  =  $\dfrac{-9}{-15}$

By cross multiplication

8 × - 15 = - 9 × k

Or , 9 k = 120

∴        k = $\dfrac{120}{9}$

i.e       k = $\dfrac{40}{3}$

Hence, The value of k for inconsistent solution is 16 and  $\dfrac{40}{3}$Answer

Answer 2

Step-by-step explanation:

8x + 5y = 9

kx + 10 y = 15

let the equation be: a¹x + b¹y +c¹ =0

and a²x + b² y + c² = 0

for inconsistent solution

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

now comparing all we get ,

$\frac{8}{k} = \frac{5}{10} = \: \frac{ - 9}{15}$

now,

$\frac{8}{k} = \frac{5}{10 } \\ = > 5k = 80 \\ = > k = \frac{80}{5 } \\ = > k \: = 16$

again,

$\frac{8}{k} \: = \frac{ - 9}{ - 15} \\ = > - 9k = - 120 \\ = > k \: = \frac{ - 120}{ - 9} \\ = > \: k = \frac{40}{3}$