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August 14th, 2017, 04:48 PM   #1
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Question 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.
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August 14th, 2017, 06:52 PM   #2
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Quote:
Originally Posted by Wisteria View Post
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.
Thanks from Denis and Wisteria
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August 14th, 2017, 11:15 PM   #3
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Try to use the two equations to find the value of x, posting your working.
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August 15th, 2017, 04:28 AM   #4
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Originally Posted by skipjack View Post
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?
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August 15th, 2017, 06:19 AM   #5
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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.
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August 15th, 2017, 09:52 PM   #6
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The third one

solution is k=1, x=1
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August 15th, 2017, 10:34 PM   #7
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Quote:
Originally Posted by Wisteria View Post
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?
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August 16th, 2017, 01:41 AM   #8
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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
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August 16th, 2017, 05:11 AM   #9
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Quote:
Originally Posted by Sequoia View Post
solution is k=1, x=1
That does not answer the question asked.
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August 17th, 2017, 05:23 PM   #10
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Quote:
Originally Posted by Wisteria View Post
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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