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Thread: Linear system problem

  1. #1
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    Linear system problem

    Hi, I am in a linear algebra class, and I don't understand how to work through this problem.

    "Consider the linear system below, where a and b represent numbers"
    2x - 3y = a
    -4x+6y = b

    (a) Find the particular values for a and b which make the system consistent.
    (b) Find the particular values for a and b which make the system inconsistent.
    (c) What relation between the values of a and b implies the system is consistent.


    When I try to solve the system with elimination, both of the x and y variables disappear on the first step. Not sure what to do.
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  2. #2
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    Re: Linear system problem

    Quote Originally Posted by MrJank View Post
    Hi, I am in a linear algebra class, and I don't understand how to work through this problem.

    "Consider the linear system below, where a and b represent numbers"
    2x - 3y = a
    -4x+6y = b

    (a) Find the particular values for a and b which make the system consistent.
    (b) Find the particular values for a and b which make the system inconsistent.
    (c) What relation between the values of a and b implies the system is consistent.
    Suppose that $(a,b)=(5,-10)$
    Now multiply the first by $2$ and add to second. Is that a consistent system? WHY?

    NOW Suppose that $(a,b)=(5,10)$
    Now multiply the first by $2$ and add to second. Is that an inconsistent system? WHY?

    YOU FINISH
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  3. #3
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    Re: Linear system problem

    Quote Originally Posted by MrJank View Post
    Ok I tried other values, and it seems that as long as b = -2a the system is consistent, am i on the right track and is there a better way to word that?

    What i mean is $(a,b) = (a, -2a)$ where b is always the opposite sign of a
    If $(a,b) = (a, -2a)$ and $a=0$ they do not have opposite signs, but the system is consistent otherwise.
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  4. #4
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    Re: Linear system problem

    Quote Originally Posted by MrJank View Post
    Thank you.
    How did you come up with (5,-10)?
    It looks like you added the coefficients together..
    Well if you try to solve a $2\times 2$ system of linear equations, you multiply on by a constant and add to the other to eliminate a variable. In this problem we must get $0=0$ So I just chose numbers that work.
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  5. #5
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    Re: Linear system problem

    YOU FINISH
    @MrJank: You've got some help. See what you can do from here. If you still can't get it then just say so.

    -Dan
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  6. #6
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    Re: Linear system problem

    Quote Originally Posted by MrJank View Post
    Hi, I am in a linear algebra class, and I don't understand how to work through this problem.

    "Consider the linear system below, where a and b represent numbers"
    2x - 3y = a
    -4x+6y = b
    Multiplying the first equation by -2, -4x+ 6y= -2a. So we have -4x+ 6y= -2a and -4x+ 6y= b. In order that the be a solution (that the equations be "consistent") we must have b= -2a. Any other values will make the system inconsistent.

    (a) Find the particular values for a and b which make the system consistent.
    (b) Find the particular values for a and b which make the system inconsistent.
    (c) What relation between the values of a and b implies the system is consistent.


    When I try to solve the system with elimination, both of the x and y variables disappear on the first step. Not sure what to do.
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  7. #7
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    Re: Linear system problem

    So let's not do elimination. Try this:

    2x - 3y = a
    -4x+6y = b ... divide each side by -2 ... so you get:

    2x - 3y = a
    2x - 3y = b/(-2)

    Now both left hand sides have the same value, no matter what x and y are. So if you subtract equation 1 from equation 2, you get:

    0 = a - b/(-2)
    or
    b = -2a ... let's call this a "condition".

    Numbers a and b that do not meet this condition make the system inconsistent.
    Numbers a and b that meet this condition make the system consistent. Which means that there is at least one solution for x and y.
    How so, you might ask. Well, if you use the condition to substitute in the second equation, you get these 2 equations:

    2x - 3y = a
    -4x+6y = -2a

    or

    2x - 3y = a
    2x - 3y = a

    So you end up with 2 identical equations. All they tell you is that: y = (2/3)x - (1/3)a. As you know, you cannot solve 2 variables from 1 equation. This one equation only tells you what is the relationship between the 2 variables (but not their concrete values).

    Later you will learn about dependence/independence of the system of linear equations.
    https://en.wikipedia.org/wiki/System...s#Independence
    Then you will be able to run a simple test on the matrix that describes your system of equations and quickly tell that the system is dependent.
    (Meaning: one equation is just a linear combination of other equations and does not really bring any new information about the relationship of x and y.)
    Last edited by troymius; Sep 3rd 2018 at 04:54 AM.
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