Question

For what value of k the system
2x+ky=1
3x-5y=7
will have
(i)unique solution
(ii)no solution.
Is there a value of k for which the system has infinitely many solutions?

Answer 1

I hope my solution will be helpful for you

Answer 2

Answer:

The correct answer will be -

(i) $k \neq \frac{-10}{3}$

(ii) $k = \frac{-10}{3} \ and \ k\neq \frac{-5}{7}$

Step-by-step explanation:

Given,

The system of equations are

2x + ky = 1

3x - 5y = 7

Now for the first Part

(I) Unique solution -

For a system of equations of the form -

a₁x + b₁y = c₁

a₂x + b₂y = c₂

For a unique solution, it must satisfy -

$\frac{a_1}{a_2}\neq \frac{b_1}{b_2}$

So from the question -

$\frac{2}{3}\neq \frac{k}{-5} \\\\k \neq \frac{-10}{3}$

So $k \neq \frac{-10}{3}$ is the answer. It can be any value other than that.

(II) No solution -

For a system of equations of the form -

a₁x + b₁y = c₁

a₂x + b₂y = c₂

For no solution, it must satisfy -

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

So, from the question -

$\frac{2}{3} =\frac{k}{-5} \neq \frac{1}{7}\\ \\k = \frac{-10}{3} \ and \ k\neq \frac{-5}{7}$

#SPJ2