# Why { $z-x-y=0$ , $z-2x=0$ , $2x+y-3z=0$ } cannot be solved this way?

I was recently solving a system of linear equations, 3 equations and 3 unknowns. I first solved via Row Reduction of the matrix and got a valid answer, but my friend attempted to solve the system using informal algebra methods and got the wrong answer. I know his answer is wrong, but I am struggling to explain what mathematical rule he broke.

Here is the system:

$$z-x-y=0$$

$$z-2x=0$$

$$2x+y-3z=0$$

Combining the first and third equation, one gets $$x=-2y$$. Plugging this back into equation one, one gets $$z=-y$$. Setting $$x=1$$, one gets the vector $$<1,-1/2,1/2>$$. This vector is valid for equations 1 and 3, but not for equation 2.

Now I know that this is not the proper technique for solving a system of three variables and that equation 2 was not used so how should one expect it to be satisfied. I know that this solution is wrong, but I am unsure how to explain what is wrong about it other than saying "that's not the way it's done." I personally made this mistake when first learning linear algebra and "that's not the way it's done" is all my teacher could say. If anyone has a better explanation for what exactly is wrong about this, I would greatly appreciate. Also, since equation 2 is not being used, it is 2 equations, 3 unknowns so there should be 2 free variables, not one (again I think this will occur if elimination is "done correctly").

• Where did $x=1$ come from? Nov 21, 2021 at 22:34
• "Setting x=1" you cannot do that because $x$ is not an independent variable (at least you have not proved that). Nov 21, 2021 at 22:36
• No: for “most” linear systems in three equations and two variables, there will be exactly one (ie theee minus two) free variable. And this is what happened here: you just forgot all about the second equation, so it’s not unexpected that the final result is one free variable for a solution of both equations (1) and (3). But there’s no reason why this solution should satisfy equation (2). Nov 21, 2021 at 22:38
• Combining the first and 3rd equations gives $x = 2z.$ And the algebra that follows is bad. Since you have 3 independent equations (and all equal to 0) the trivial solution is the only solution. Nov 21, 2021 at 22:44
• Your friend found a solution to a system of two equations. Why is he surprised that it is not a solution to a third, unused and unrelated equation? Nov 22, 2021 at 7:27

Your friend used just two equations, which (since they were linearly independent) is enough to limit the solution space to one dimension. The correct conclusion is not that $$\langle x,y,z\rangle = \langle 1,-\frac12,\frac12\rangle$$ is the solution, but that the solution has the form $$\langle x,y,z\rangle = \left\langle t,-\frac12t,\frac12t\right\rangle$$ for some real number $$t.$$

This is true. The correct solution does have that form.

Now compare this with your solution and see if you can tell what value(s) of $$t$$ give a solution to all three equations.

You wrote, correctly, that the second equation was not used at all. And the solutions of the system which consists of the other two equations are the triplets $$(x,y,z)$$ such that $$x=-2y$$ and that $$z=-y$$ indeed. Finally, there is not reason at all to pick $$x=1$$.

• worth pointing out, adding the first and third equations leads to $x-2z=0$ rather than $x-2y=0$ as the OP thinks. Nov 21, 2021 at 22:44
• @WillJagy In fact, but adding $3$ times the first equation to the third one leads indeed to $x=-2y$. Nov 21, 2021 at 22:46

Let me present an extreme version of your friend's argument:

By the first equation, we have

$$z = x + y$$

Then, simply choose $$x = y = 1$$ and conclude $$z = 2$$.

Thus, $$(1,1,2)$$ is a solution to the system!

Or to a ridiculous degree:

By ignoring all equations, we can choose $$x = y = z = 1$$ and so $$(1,1,1)$$ is a solution!

I hope this illuminates the issue with your friend's solution. Each equation potentially restricts the possible solutions to the system, and so each equation not used potentially un-restricts those solutions to ones which are not actually valid.

One more example to demonstrate this fact:

Consider the system of two equations for one variable, \begin{align*} x &= 1 \\ x &= 2 \end{align*} By limiting our scope to only one equation, we may conclude either that $$1$$ is a solution or that $$2$$ is a solution, but in reality there are (obviously) no solutions, and so both are wrong.

We can begin by substituting $$(2)$$ into $$(1)$$ or $$(3)$$. The problem is, using $$(2)$$, we find a contradiction between $$(4)$$ and $$(6)$$

\begin{align} z-x-y=0 \implies -x-y+z&=0 \tag{1}\\ z-2x=0 \implies z&=2x \tag{2}\\ 2x+y-3z=0\implies 2(x)+y-3z&=0 \tag{3} \end{align} Substitute $$(2)$$ into $$(1)$$ then $$(4)$$ into $$(3)$$

\begin{align} -x-y+(2x)=0\implies x&=y \tag{4}\\ 2(x)+y-3z=0\implies 3y-3z&= 0\tag{5}\\ \implies y=z\implies y&=2x \tag{6} \end{align}

or we can combine $$(1)$$ and $$(3)$$ as you did.

Sometimes rearranged things are earier to "see" \begin{align} z-x-y=0 \implies -x-y+z=0 \tag{1}\\ z-2x=0 \implies -2x+0y+z=0 \tag{2}\\ 2x+y-3z=0\implies 2(x)+y-3z=0 \tag{3} \end{align} Add $$(1)$$ to $$(3)$$, substitute $$(4)$$ into $$(3)$$ then $$(5)$$ into $$(6)$$, etc.

\begin{align} x-2z=0\implies x&=(2z) \tag{4} \\ \implies z&=\frac{x}{2} \tag{5}\\ 2(2z)+y-3z=0\implies y&=(-z) \tag{6}\\ \implies y&=\frac{-x}{2} \tag{7} \end{align}

Setting $$x=1\quad \bigg(x,\dfrac{-x}{2},\dfrac{x}{2}\bigg) =\bigg(1,\dfrac{-1}{2},\dfrac{1}{2}\bigg)$$

It might be useful to see what you actually did solve: the set of three equations:

$$z-x-y=0$$

$$x=1$$

$$2x+y-3z=0$$

I've removed the equation that wasn't used, and replaced it with an equation that was used. <1,−1/2,1/2> is a solution to this particular problem. For this particular problem, I wouldn't start by combining 1 and 3. Instead, I'd first substitute 2 into 1 and then 2 into 3.

It's still a similar sort of problem: the intersection of three planes. But if should be fairly clear that by choosing different planes, you can get different solutions.