Find the value of k for which the system of equations
$k x - y = 2$
$6 x - 2 y = 3$
has a unique solution?

Open in App

Solution

The given system of equations is
$k x - y - 2 = 0$
$6 x - 2 y - 3 = 0$

It is of the form $a_{1} x + b_{1} y + c_{1} = 0$
$a_{2} x + b_{2} y + c_{2} = 0$ ,

where $a_{1} = k, b_{1} = - 1, c_{1} = - 2$ and $a_{2} = 6, b_{2} = - 2$ and $c_{2} = - 3$

(i) The given system will have a unique solution, if

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ i.e, if $\frac{k}{6} \neq \frac{- 1}{- 2}$ i.e., $k \neq 3$

So, the given of equations system will have a unique solution, if $k \neq 3$ .

Suggest Corrections

Similar questions

k \neq m

k x - y = 2

6 x - 2 y = 3

k

k x - y = 2

6 x - 2 y = 3

k

k x + 2 y = 5

3 x + y = 1

Find the value for k which each of the following systems of equations has a unique solution:

$k x + 3 y = (k - 3), 12 x + k y = k .$

Join BYJU'S Learning Program