Write down values of $a$ and $b$ for which this system of equations has a non unique solution As a part of a task I got the values of $x$, $y$ and $z$ by solving a system of three equations. The values of $x$, $y$ and $z$ are as follows:
$x=\frac {5-2a-3b}{2b}$
$y=\frac {4a+11b-10}{2b}$
$z=\frac {2a-5}{2b}$
Now I have to find the values of a and b such that the system of equations has a non unique solution.
I have no idea how to proceed here. I understand that by having a non unique solution, the three planes must intersect in a line. Any hints?
The answer at the back of the book is: $a=\frac52$ and $b=0$
Full question:
Consider the following system of equations:
$3x+y+z=1$
$x+y-z=4$
$2x+y+bz=a$
where $a$ and $b$ are constants.
Solve the system in terms of $a$ and $b$
Hence, write down the values of $a$ and $b$ for which this system of equations has a non unique solution and state its geometric meaning. 
 A: Your solution to the problem is partially correct. The $x$, $y$ and $z$ that you solved for are indeed a solution to the system, if $b\not= 0$. So with you technique of solving the system you 'lost' the scenario when $b=0$. 
So given you did not yet learn anything about matrices and maybe about linear independence and that kind of stuff, a solution technique might be the following: 


*

*You start solving the system in the way you apparently did: by substitution

*At some point you have to divide by $b$. From that point on you have to do a case distinction

*First you finish solving the system as if $b\not= 0$

*When done, go back to situation before you divided by $b$, and now fill in $b=0$ for all $b$'s, and solve the resulting system.


You'll notice that in step $4$ eventually $x$,$y$ and $z$ all cancel out in the last equation, and you are left with $a=\frac{5}{2}$. And at that point you'll have you non-unique solution, which looks something like
$$x=x$$
$$y=f(x)$$
$$z=g(x,y)$$
(Step 3 can be skipped if you aren't interested in the case $b\not=0$, and of course the order of steps $3$ and $4$ can be changed)

Edit: oke here are the details (note there are other ways of doing this, this is just one)
Start by saying (from the second equality)
$$y=4+z-x\ \ \ \ \ \ \ (1)$$
So that by substitution in the first equality
$$x=-z-\frac{3}{2}\ \ \ \ \ (2)$$
And by substitution in the third equality
$$x+4+(b+1)z=a\ \ \ \ \ \ (3)$$
So now filling in $(2)$ in $(3)$ we find
$$-z-\frac{3}{2}+4(b+1)z=\frac{5}{2}+bz=a$$
So
$$z=\frac{a-\frac{5}{2}}{b}$$
And here we say wait. We just divided by $b$, is the allowed? Well it is if $b\not=0$. If we now proceed from here as if indeed $b\not=0$, then we find the solution that you listed in your post.
Now let's see what happens when we say instead that $b=0$. Then we are back in the situation 
$$\frac{5}{2}+bz=a$$
And since now $b=0$, this reduces to
$$a=\frac{5}{2}$$
This is what I meant we I said at we obtain a relation from with $x,y$ and $z$ are canceled out. 
So we find
$$z=z$$
$$x=-z-\frac{3}{2}$$
$$y=4+z-x$$
As a whole family of solution that are all valid under the assumption that $b=0$ and $a=\frac{5}{2}$.
A: If you solve the first two equations for $x,y$ you get
$x = -{3 \over 2} -z$, $y={11 \over 2} +2 z$.
Then the last equation becomes
$bz + {5 \over 2} = a$.
So, if $b=0$ and $a \neq {5 \over 2}$ there is no solution. Otherwise there is a solution which is $z = {a-{5 \over2} \over b}$ (and the values of $x,y$ follow
from the above formulae).
A: Your book is incorrect on a subtle but important logical point.
The equation $x = \frac 00$ does not necessarily have an infinite number of solutions.  $\frac 00$ is undefined, so to say that (for example) $x = 5$ is a solution to $x=\frac 00$ is wrong.  
On the other hand, $0\cdot x = 0$ does have an infinite number of solutions.  Any number makes that statement true.
It would have been better to ask about
$$2b \cdot x= 5-2a-3b$$
$$2b \cdot y=4a+11b-10$$
$$2b \cdot z=2a-5$$
Here you can see that to convert them all to a form $0 \cdot x = 0$, you must choose $b = 0$:
$$0 \cdot x= 5-2a$$
$$0 \cdot y=4a-10$$
$$0 \cdot z=2a-5$$
And there you can see that choosing $a = \frac 52$ makes all the equations into
$$0 \cdot x= 0$$
$$0 \cdot y= 0$$
$$0 \cdot z= 0$$
Which does have an infinite number of solutions.
