Question

find the value of k for which the following system of equations has a unique solutions 1 . kx +2y= 5 , 3x+y=1

Answer 1

Answer:

The value of k is all number except $k=6$

Step-by-step explanation:

Given : Equations $kx+2y=5,3x+y=1$

To find : The value of k for which the following system of equation has a unique solution.

Solution :  

When the system of equation is in form $ax+by+c=0, dx+ey+f=0$ then the condition for a unique solutions is  

$\frac{a}{d}\neq\frac{b}{e}$

Comparing, a=k , b=2 , c=-5, d=3, e=1, f=-1

Substituting the values,

$\frac{k}{3}\neq\frac{2}{1}$

Cross multiply,

$k\times 1\neq 2\times 3$

$k\neq 6$

Therefore, The value of k is all number except $k=6$

Answer 2

Answer:

$For \:all\:real \:values \:of\\k ,the \: equations\:shows \\unique \: solution\:except \\k≠6$

Step-by-step explanation:

$Given,\: system\:of\: equations:\\kx+2y-5=0--(1)\\3x+y-1=0$

$Compare\:these \: equations\\with \:a_{1}x+b_{1}y+c_{1}=0\\a_{2}x+b_{2}y+c_{2}=0$

$a_{1}=k,\:b_{1}=2,\:c_{1}=-5\\a_{2}=3,\:b_{2}=1,\:c_{2}=-1$

$\frac{a_{1}}{a_{2}}≠\frac{b_{1}}{b_{2}}\\(Given \: equations \:have \: unique\: solution)$

$\implies \frac{k}{3}≠\frac{2}{1}$

$\implies k\times 1≠2\times 3$

$\implies k ≠ 6$

Therefore,

$For \:all\:real \:values \:of\\k ,the \: equations\:shows \\unique \: solution\:except \\k≠6$