Question

Find the value of k for which the system of equations x-2y=1 and 2x-ky=2 has infinitely many solutions. *

Answer 1

Step-by-step explanation:

a1/a2=b1/b2=c1/c2

this is the formulae. for co incident line and have infinately solutions

k=4

because

1/2=2/4=1/2then

1/2=1/2=1/2 wiĺ come

i hope it more helps.you

Answer 2

Answer:

The value of k for which the system of equations x-2y=1 and 2x-ky=2 has infinitely many solutions is 4.

Step-by-step explanation:

Equations of given lines are,

$x-2y =1$ ⇒$x-2y-1 =0$

$2x-ky=2$⇒$2x-ky-2=0$

We are given that the given lines has infinitely many solutions. That means the given lines are coincident lines.

If the lines $a_{1} x+b_{1} y+c_{1} =0$  and $a_{2} x+b_{2} y+c_{2} =0$  are coincident lines. Then,

$\frac{a_{1} }{a_{2} } =\frac{b_{1} }{b_{2} } =\frac{c_{1} }{c_{2} }$

Therefore,

$\frac{1}{2} =\frac{-2}{-k} =\frac{-1}{-2}$

Equate the first two terms and simplify it.

$\frac{1}{2} =\frac{-2}{-k}$

On cross multiplication. We get,

$-k(1)=-2(2)$

Perform multiplication on both sides to get the required answer

$k=4$

Therefore, the value of k is 4.