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 August 14th, 2017, 03:48 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, 05:52 PM   #2
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 Originally Posted by Wisteria 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.
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, 10:15 PM #3 Global Moderator   Joined: Dec 2006 Posts: 19,722 Thanks: 1807 Try to use the two equations to find the value of x, posting your working.
August 15th, 2017, 03:28 AM   #4
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 Originally Posted by skipjack Try to use the two equations to find the value of x, posting your working.
ok so I solve for x first instead of k?

 August 15th, 2017, 05:19 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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, 08:52 PM #6 Newbie   Joined: Aug 2017 From: America Posts: 5 Thanks: 0 The third one solution is k=1, x=1
August 15th, 2017, 09:34 PM   #7
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 Originally Posted by Wisteria ok so I solve for x first instead of k?
For the given equations, that's a good idea (because the working is easier than solving for y). Can you try it and post your working?

 August 16th, 2017, 12:41 AM #8 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 878 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, 04:11 AM   #9
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 Originally Posted by Sequoia solution is k=1, x=1

August 17th, 2017, 04:23 PM   #10
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Quote:
 Originally Posted by Wisteria 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.
From the first equation, x= k- 2ky. So 2kx+ 4ky= 4k- 2 can be written as $\displaystyle 2k(k- 2ky)+ 4ky= 2k^2- (4k^2- 4k)y= 4k- 2$. Them $\displaystyle (8k- 4k^2)y= 2k^2- 4k+ 2$.

$\displaystyle y= \frac{2k^2- 4k+ 2}{8k- 4k^2}$

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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