Question

2x+3y-5=0
6x+ky-15=0
find 'k'
has infinity many solutions

Answer 1

Answer:

k=9

Step-by-step explanation

For any equation to have infinitely many solutions, the following condition is required-

a1/a2 = b1/b2 =c1/c2

Where, a1=2 ; a2=6; b1=3; b2= k; c1=-5; c2=-15

Therefore, ATQ

=> 2/6=3/k

=>1/3=3/k

=>k=9

hence solved

Answer 2

The value of k is 9.

Step-by-step explanation:

The given system of equations are

$2x+3y-5=0$

$6x+ky-15=0$

If two equations $a_1x+b_1y+c_1=0$ and  $a_2x+b_2y+c_2=0$ has infinity many solutions, then

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

Given equations has infinity many solutions, then

$\dfrac{2}{6}=\dfrac{3}{k}=\dfrac{-5}{-15}$

$\dfrac{1}{3}=\dfrac{3}{k}=\dfrac{1}{3}$

$\dfrac{1}{3}=\dfrac{3}{k}$

On cross multiplication we get

$k=9$

Therefore, the value of k is 9.

#Learn more

Find k for which the system 2x + 3y -5=0, 4x + ky -10=0 has infinitely many solutions.

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