\begin{align} &{-}2y+2z-1=0 \tag{4} \\[4px] &{-}2x+4y-2z-2=0 \tag{5} \\[4px] &\phantom{-2}x-y+3/2=0 \tag{6} \end{align}

Equation (6) is the sum of (4) and (5). There are only two independent equations. Putting $z=0$ in (5) and (6) and solving for x and y, we have \begin{align} x&=-2 \\[4px] y&=-1/2 \end{align}

  • equation (6) is the sum of (4) and (5): OK, I see it

  • There are only two independent equations: I didn't get; what does this sentence mean?

  • Putting $z=0$ in (5) and (6): why putting z=0 in equation (5) and (6)?

Please help


Any value given to $x$, $y$ and $z$ that satisfy equations (4) and (5) will also satisfy equation (6). So the last equation gives no new information.

The fact that $z$ disappears if we sum (4) and (5) means that it is “free”: we can assign it any value and determine suitable values for $x$ and $y$ that satisfy the equations.

Using $z=0$ will give one solution for the system, but it has infinitely many solutions, one for each value given to $z$.

For instance, if we use $z=1$, we get $x=-1$ and $y=1/2$.

The general solution is \begin{cases} x=-2+z \\[4px] y=-\dfrac{1}{2}+z \end{cases}

  • $\begingroup$ Thank you, now my doubts are clear $\endgroup$ – Akash Dec 13 '18 at 13:28

(This is too long for a comment so I post it as a solution).

In order to solve any equation with multiple variables, a good way to approach is to think how much "information" each equation gives.

We see the first equation, so it gives us one information. The second gives us a second information, because it cannot be derived from previous information.

But looking at the third equation (or number 6), we see that you can get this result by adding the two previous equations together. Therefore, it does not give us any new information. Now we have three unknowns and two equations that give us information about them. These two equations are said to be independent because they give us information about the variables. The third equation is a linear combination of the other two, so it's NOT independent of the two others.

Because we have two equations and three unknowns, we have an infinite amount of solutions. Apparently, in the example, we are interested in obtaining at least one of them. Due to the linearity of the equations, we we are free to set any value to any of the variables, and we will obtain the other two. In this case, the author has decided to set $z=0$ in order to get a solution.

  • $\begingroup$ Thank you i got it $\endgroup$ – Akash Dec 13 '18 at 13:29

You can also think of as follows.

  • A single variable linear equation represents a single point.
  • A two-variable linear equation represents a line. If there are two two-variable linear equations, their solution is the intersection point of them as long as both equations are independent.

  • A three-variable linear equation represent a plane. If there are 3 three-variable equations, their solution is the intersection point of them as long as they are independent.

So if you have two 3-variable linear equations (or three 3-variable linear equations in which two of them are dependent), you cannot get a single point of intersection but a collection of points which are on a line, for example, $x+y+5/2=0$ in your case.


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