# Thread: Solving system of linear equations

1. ## Solving system of linear equations

Find the parameter $\displaystyle m$ in order that the system of linear equations:

$\displaystyle x+y+z=2$
$\displaystyle 2x+3y+2z=5$
$\displaystyle 2x+3y+(2m-1)z=2m+1$

a) has one unique solution
b) has no solutions
c) has infinitely many solutions

Thank you

2. Okay, what have you done? What happens if you try to solve x, y, and z? I suspect that you will get x, y, or z, whichever you solve for first, equal to a fraction. There will be one solution if the denominator is not 0, no solution if the denominator is 0 but the numerator is not, and an infinite number of solutions if both numerator and denominator are 0 . What values of a make each of those things happen?

(Look closely at the last two equations. Tthere is an obvious first step that will simplify things very fast!)

3. Hi HallsofIvy,

Using matrices, I came to an equivalent system of linear equations and found that for m=3/2 the system is impossible. For any other value other than m=3/2 the equation has an unique solution. (I don't know if this statement is completely right, and if it's an answer to the first two questions).

I don't know how to find a value of m in order that the system can have infinitely many solutions.

If you can write me the complete solution and required information, I would be really grateful.

Thank you.

4. Then you have done very well.

If you subtract the second equation from the third, both x and y are eliminated and you have
$\displaystyle (2m- 3)z= 2m- 4$

If m 2m- 3 is NOT 0, that is, if m is NOT 3/2, then you can solve for z by dividing- if m is not 3/2, then there exist a single solution for z. If m= 3/2, the the equation becomes 0z= 3- 4= -1 which is impossible. There is no solution.

Because there is no m in either of the first two equations, if m is not 3/2, we can put z= (2m-4)/(2m-3) into the other two equations and find no new restriction on x and y. There is NO value of m for which there are an infinite number of solutions.

(If the equations had reduced to, say, (m-2)(m+1)z= m+1, then we could say that, for m= 2, we get 0z= 3 which is impossible but for m=-1, it reduces to 0z= 0 which is true for all z. That would give an infinite number of solutions but that does not happen in the given problem.)

5. Thank you very much for confirmation.