Question

it the following system equations have infinity many solutions then value of K will be:2x+3y=5and4xky=10​

Answer 1

Answer:

k=6

Step-by-step explanation:

2x+3y=5

2x+3y=54x+ky=10

2x+3y=54x+ky=10For infinitely many solutions:-

2x+3y=54x+ky=10For infinitely many solutions:-a1/a2 = b1/b2 = c1/c2

2x+3y=54x+ky=10For infinitely many solutions:-a1/a2 = b1/b2 = c1/c2a1=2, a2=4 ; b1=3 , b2=k; c1=5 , c2=10

2x+3y=54x+ky=10For infinitely many solutions:-a1/a2 = b1/b2 = c1/c2a1=2, a2=4 ; b1=3 , b2=k; c1=5 , c2=10so, 2/4 = 3/k = 5/ 10

2x+3y=54x+ky=10For infinitely many solutions:-a1/a2 = b1/b2 = c1/c2a1=2, a2=4 ; b1=3 , b2=k; c1=5 , c2=10so, 2/4 = 3/k = 5/ 10hence, 1/2 = 3/k

2x+3y=54x+ky=10For infinitely many solutions:-a1/a2 = b1/b2 = c1/c2a1=2, a2=4 ; b1=3 , b2=k; c1=5 , c2=10so, 2/4 = 3/k = 5/ 10hence, 1/2 = 3/kBy cross multiplication:-

2x+3y=54x+ky=10For infinitely many solutions:-a1/a2 = b1/b2 = c1/c2a1=2, a2=4 ; b1=3 , b2=k; c1=5 , c2=10so, 2/4 = 3/k = 5/ 10hence, 1/2 = 3/kBy cross multiplication:-We get, K=6

Answer 2

Given equations are 2x+3y−5=0 and 4x+ky−10=0

Then 2x+3y−5=0⇒2x+3y=5

4x+ky−10⇒4x+ky=10

∴

4

2

=

k

3

=

10

5

.... [Since, they have an infinite number of solutions]

⇒

2

1

=

k

3

=

2

1

⇒

2

1

=

k

3

⇒k=6