Question

Find the value of k for which each of the following pairs of equation have infinite many solution :
2x + (k-2)y = k
6x + ( 2k -1)y= 2k +5

Answer 1

Answer:

Let A={1,2,3,4 } and B=N , let f:A → B be defined by f(x)= x2 then

i. Find the range of f (ii)Identify the type of function

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

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

have infinite number of solutions if $\displaystyle \: \: \frac{a_1}{a_2} = \frac{b_1}{b_2}=\frac{c_1}{c_2}$

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

Given pair of linear equations

$2x +(k- 2) y = k \: \: and \: \: 6x + (2k-1)y =2k+5$

Comparing with

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

We get

$\displaystyle \: a_1 = 2 \: , \: b_1 = k- 2 \: , c_1= - k \: and \: \: a_2 = 6 \: , \: b_2 = 2k -1\: , \: \: c_2=-(2k+5)$

So the given pair of Straight Lines have infinite number of solutions if

$\displaystyle \: \frac{2}{6} = \frac{k - 2}{2k - 1} = \frac{k}{2k + 5}$

Now

$\displaystyle \: \frac{2}{6} = \frac{k - 2}{2k - 1} \: \: gives$

$4k - 2 = 6k - 12$

$\implies \: 2k = 10$

$\implies \: k \: = 5$

Again

$\displaystyle \: \frac{2}{6} = \frac{k}{2k + 5} \: \: \: \: \: gives$

$\implies \:6k =4 k + 10$

$\implies \: 2k = 10$

$\implies \: k = 5$

So the required value of k is 5