Question

For what value of k’ given equation has no solution

x– 4y = 6 & 3x + ky = 5​

Answer 1

Answer:

if the equation has no solution then,

a1/a2=b1//b2

1/3= -4/k

1/3 k = -4

k= -4 x 3 = -12

therfore k= -12

Step-by-step explanation:

Answer 2

$\huge\boxed{\underline{\underline{\green{Solution}}}} </p><p>$

$\displaystyle \: \longmapsto \: \:$FORMULA TO BE IMPLEMENTED :

A pair of Straight Lines

$\displaystyle \: a_1x+b_1y+c_1=0 \: and \: \: a_2x+b_2y+c_2=0$

Is said to have no solution if $\displaystyle \: \: \frac{a_1}{a_2} = \frac{b_1}{b_2}$

$\displaystyle \: \longmapsto \: \:$CALCULATION :

Given pair of linear equations

$x - 4y = 6 \: \: and \: \: 3x + ky =5$

Comparing with

$\displaystyle \: a_1x+b_1y+c_1=0 \: and \: \: a_2x+b_2y+c_2=0$

We get

$\displaystyle \: a_1 = 1 \: , \: b_1 = - 4 \: , c_1= -6 \: and \: \: a_2 = 3 \: , \: b_2 = k\: , \: \: c_2= - 5$

So the given two equations have no solution if

$\displaystyle \: \: \frac{a_1}{a_2} = \frac{b_1}{b_2}$

$\implies \: \displaystyle \: \frac{1}{3} = \frac{ - 4}{k}$

$\therefore \: k \: = - 12$