Question

Find the value of k for which each of the following systems of equations have infinitely many solution:
2x+3y-5=06x+ky-15=0

Answer 1

k = 9

Step-by-step explanation:

Given:  

2x +3y -5 = 0

6x +ky -15 = 0

The system of equations has infinitely many solutions.

a1 = 2, b1 = 3, c1 = -5

a2 = 6, b2 = k, c2 = -15

So  a1/a2 = b1/b2 = c1/c2

2/6 = 3/k

Therefore k = 9

Answer 2

$\Large{\underline{\underline{\bf{Solution :}}}}$

As, it is given equations has infinite many solutions.

We know the case of infinite many solutions.

$\Large{\star{\boxed{\rm{\frac{a_{1}} {a_{2}} = \frac{b_1}{b_2} = \frac{c_1}{c_2}}}}}$

Where,

a1 = 2, a2 = 6

b1 = 3, b2 = k

c1 = -5, c2 = -15

________________[Put Values]

$\sf{→\frac{2}{6} = \frac{3}{k} = \frac{-5}{-15}} \\ \\ \sf{→\frac{2}{6} = \frac{3}{k}} \\ \\ \sf{→k = \frac{\cancel{6} \times 3}{\cancel{2}}} \\ \\ \sf{→k = 3 \times 3}\\ \\ \sf{→k = 9} \\ \\ \Large{\star{\boxed{\sf{k = 9 }}}}$