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 Find the value of m where $k$, for which the system of equations $kx−y=2$ and $6x−2y=3$ has a unique solution.
Given: Equations $kx−y=2$ and $6x−2y=3$
To do: To find the value of m where $k$ for the unique solution of given system of equations.
Solution:
Given system of equations are
$6x - 2y = 3$
$6x - 2y - 3 = 0 ----( 1 )
$kx - y = 2$
$kx - y - 2 = 0 ----( 2 )$
Compare above equations with
$a1 x + b1 y + c1 = 0$ and
$a2 x + b2 y + c2 = 0$ , we get
$a_1=6,\ b_1=-2,\ c_1=-3$ ;
$a_2=k,\ b_2=-1,\ c_2=-2$ ;
Now ,
$\frac{a_1}{a_2}\
eq \frac{b_1}{b_2}$ [ Given they have Unique solution ]
eq \frac{b_1}{b_2}$ [ Given they have Unique solution ]
$\Rightarrow \frac{6}{k}\
eq\frac{( -2)}{( -1)}$
eq\frac{( -2)}{( -1)}$
$\Rightarrow \frac{6}{k}\
eq{2}$
eq{2}$
$\Rightarrow \frac{k}{6}\
eq\frac{1}{2}$
eq\frac{1}{2}$
$\Rightarrow k\
eq \frac{6}{2}$
eq \frac{6}{2}$
$\Rightarrow k\
eq 3$
eq 3$
Therefore , For all real values of $k$ , except $k\
eq3$, given system of equations have unique solution.
eq3$, given system of equations have unique solution.
Updated on: 10-Oct-2022
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