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$x+3y-z=2$

$x+y-z=0$

$3x+2y-3z=-1$

This is my work so far: $x+3y-z=2 -(x+y-z=0)$ which equals to $2y=2$ and therefore, $y=1$

I plugged the value of y back into the first two-equation and got $x-z=-1$ and $x-z=1$

which is infinitely many solutions, but the question asks for the precise answer. I just solved it for x and got $x=z-1$

SO my ordered triple would be $(z-1,1,??)$ I don't know the last one. I'm not sure how I'm supposed to get something in terms of z. If I plug $x=z-1$ back into the first equation, z just gets eliminated. How do I find z?

QuantumPi
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2 Answers2

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If you plug $y=1$ in the second equation, you would also get $x-z=-y=-1$. This happens as well for the third one: $$3x-3z=-1-2y=-1-2=-3\implies x-z=-1.$$ Therefore, the solution to this linear system has the form $$\{(z-1,1,z)\mid\text{$z$ is a constant}\}.$$ That is, this system is under-determined and has infinitely many solutions.

Bernard Pan
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  • wait but why is the last one just z? – QuantumPi Mar 28 '21 at 15:15
  • You regard it as an arbitrary constant and express $x$ as some functions of $z$. Its value cannot be determined in this linear system. – Bernard Pan Mar 28 '21 at 15:16
  • ohhhhhhhhhhh i get it now. So, if I had solved this differently in terms of y, I would just put y in the middle of the ordered pair right? – QuantumPi Mar 28 '21 at 15:17
  • No. In this system, $y$ is uniquely determined as a constant $1$, but the values of $x$ and $z$ are undetermined. Nonetheless, $x$ and $z$ are found to satisfy a relation $x-z=-1$. Thus we can regard $z$ as a constant and express $x$ as a function of $z$: $x=z-1$; or alternatively, regard $x$ as a constant and express $z$ as a function of $x$: $z=x+1$. In the latter case, the solution is expressed as ${(x,1,x+1)\mid\text{$x$ is a constant}}$. – Bernard Pan Mar 28 '21 at 15:21
  • ohhhh right. Thanks for the clarificiation. – QuantumPi Mar 28 '21 at 15:21
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You can say "let $x=k\in\mathbb{R}$", meaning that we let $x$ be any number in $\mathbb{R}$ and then $z=k+1$. So the solutions are $(k,1,k+1)$ where $k\in\mathbb{R}$. The same thing is to say "let $z=k\in\mathbb{R}$" and then the solutions we would have are of the form $(k-1,1,k)$ which you can check form the same set of solutions with the previous set of solutions. The idea is that every pair of $x,z$ that satisfies the equation $x-z=-1$ gives a solution without having any other limitation upon the values of $x$ and $z$.

Just_a_student
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