August 14th, 2017, 04:48 PM  #1 
Newbie Joined: Aug 2017 From: Fiji Posts: 2 Thanks: 0 Math Focus: calculus  I need help solving this
How should we actually solve for : Find the value(s) of the constant k such that the system of linear equations: x +2ky = k, 2kx +4ky = 4k−2 has (i) No solution. (ii) An inﬁnite number of solutions. (iii) Exactly one solution. I am confused. 
August 14th, 2017, 06:52 PM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,461 Thanks: 489 Math Focus: Yet to find out.  I think first you should review what each of i, ii, and iii mean. Maybe even write out their definitions on paper, then go back to the system of equations and investigate values of k that might satisfy the definitions.

August 14th, 2017, 11:15 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,250 Thanks: 1439 
Try to use the two equations to find the value of x, posting your working.

August 15th, 2017, 04:28 AM  #4 
Newbie Joined: Aug 2017 From: Fiji Posts: 2 Thanks: 0 Math Focus: calculus  
August 15th, 2017, 06:19 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,879 Thanks: 766 
From the first equation, x= k 2ky. Put that into the second equation and solve for y. You should get y equal to a fraction depending on k: y= f(k)/g(k). There will be one unique solution as long as g(k) is not 0. There will be no solution if g(k)= 0 and f(k) is not 0. There will be an infinite number of solutions if g(k)= 0 and f(k)= 0.

August 15th, 2017, 09:52 PM  #6 
Newbie Joined: Aug 2017 From: America Posts: 5 Thanks: 0  The third one
solution is k=1, x=1

August 15th, 2017, 10:34 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 18,250 Thanks: 1439  
August 16th, 2017, 01:41 AM  #8 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 872 Thanks: 60 Math Focus: सामान्य गणित 
Graphically, (i) No solution = lines are parallel but not overlapping (ii) An inﬁnite number of solutions = lines are overlapping (iii) Exactly one solution = lines are intersecting 
August 16th, 2017, 05:11 AM  #9 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,879 Thanks: 766  
August 17th, 2017, 05:23 PM  #10  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,879 Thanks: 766  Quote:
$\displaystyle y= \frac{2k^2 4k+ 2}{8k 4k^2}$ There is one solution as long as $\displaystyle 8k 4k^2$ is not 0. There is no solution if $\displaystyle 8k 4k^2= 0$ and $\displaystyle 2k^2 4k+ 2$ is not 0. There are an infinite number of solutions if both are 0.  

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