# Thread: Linear system problem

1. ## 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.

2. ## Re: Linear system problem

Originally Posted by MrJank
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

3. ## Re: Linear system problem

Originally Posted by MrJank
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.

4. ## Re: Linear system problem

Originally Posted by MrJank
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.

5. ## 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

6. ## Re: Linear system problem

Originally Posted by MrJank
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.

7. ## 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.)