Math, asked by madhur265gmailcom, 11 months ago

Find the value of k for which the following system of equations has no solution:
2 x - ky + 3 = 0 and 3x +2y-1 =0​

Answers

Answered by tjlngpl2503
4

Answer:

-4/3 = k

Step-by-step explanation:

2x-ky+3=0

3x+2y-1=0

On comparing the equations wit their standard forms

we get ,

a1=2 , b1=-k , c1=3 , a2=3 , b2=2 , c2=-1

For no solutions

a1/a2 = b1/b2 not= c1/c2

now put the values

2/3 = -k/2 not= 3/-1

from a1/a2 = b1/b2

2/3 = -k/2

on taking 1/2 from RHS to LHS and then solving we get

4/3 = -k

-4/3 = k

(only if the school says to then do the rest)

if you have to show not= , then,

from b1/b2 not= c1/c2

-k/2 not=3/-1

on taking 1/2 from LHS to RHS and then solving we get

-k not= 6/-1

-k not= -6

k not= 6

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