Linear System Solutions for a Unique Value of a | Math 115 Homework

AI Thread Summary
The linear system presented has varying solutions based on the value of 'a'. For a unique solution, 'a' must not equal 2, as this leads to parallel lines with no intersection. When 'a' equals 2, the system has no solutions due to the lines being parallel but having different y-intercepts. The case for infinitely many solutions occurs when the equations represent the same line, which does not happen for any value of 'a' in this scenario. The general solution for a = 1 was correctly found as (2/3, 1/3).
JettyZ
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Homework Statement


I'm trying to solve #1 here:
http://www.student.math.uwaterloo.ca/~math115/Exams/M115.FE.pdf

The problem is:
Consider the linear system:
x + ay = 1
ax + 4y = 2

(a) For what values of a does the system have a unique solution?
(b) For what values of a does the system have infinitely many solutions?
(c) For what values of a does the system have no solution?
(d) For a = 1 find the general solution.


The Attempt at a Solution


(a) At a=2, the slope is the same. I figure any value but 2 gives unique solutions. But I don't know if this is right.
(b) At a=2, the slopes are the same, but lines are different. I cannot find any value of a that would give the same slope and same outputs for inputs of x. DNE is the answer?
(c) At a=2, there are no solutions for the system because the lines are parallel to each other. a=2 is the answer.
(d) I think this part was done correctly. I used substitution and solved for the x and y values at a=1 and resulted with (2/3, 1/3)

I would like to know about a, b, and c. I don't know if I did these right.

Thanks.
 
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Ignoring (a) and (b) for the moment:
JettyZ said:
(a) At a=2, the slope is the same. I figure any value but 2 gives unique solutions. But I don't know if this is right.
(b) At a=2, the slopes are the same, but lines are different. I cannot find any value of a that would give the same slope and same outputs for inputs of x. DNE is the answer?
(c) At a=2, there are no solutions for the system because the lines are parallel to each other. a=2 is the answer.
(d) I think this part was done correctly. I used substitution and solved for the x and y values at a=1 and resulted with (2/3, 1/3)
(d) is right, but (c) is wrong. At a = 2, you have this system:
x + 2y = 1
2x + 4y = 2
Parallel lines have the same slope, but different y-intercepts. Is that the case here?
 
Hey JettyZ.

For c), there are many solutions corresponding to x = t and y = 1 - 2t for any value of t.
For a) and b) you need to use determinants and check for possibilities of inconsistent solutions.
 
b) At a=2, the slopes are the same, but lines are different. I cannot find any value of a that would give the same slope and same outputs for inputs of x.
If a= 2, your equations are x+ 2y= 1 and 2x+ 4y= 2. If you multiply the first equation by 2, what happens? What does that tell you?
 
JettyZ said:

Homework Statement


I'm trying to solve #1 here:
http://www.student.math.uwaterloo.ca/~math115/Exams/M115.FE.pdf

The problem is:



The Attempt at a Solution


(a) At a=2, the slope is the same. I figure any value but 2 gives unique solutions. But I don't know if this is right.
(b) At a=2, the slopes are the same, but lines are different. I cannot find any value of a that would give the same slope and same outputs for inputs of x. DNE is the answer?
(c) At a=2, there are no solutions for the system because the lines are parallel to each other. a=2 is the answer.
(d) I think this part was done correctly. I used substitution and solved for the x and y values at a=1 and resulted with (2/3, 1/3)

I would like to know about a, b, and c. I don't know if I did these right.

Thanks.

Do you know about determinants and their relationship to such questions? If so, use a determinant; you will see that there is a critical value of 'a' that you have missed.

RGV
 
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I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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