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I'm studying High School Algebra and it had this question:

Solve the system by equations:

\begin{align*} x + y - z &= \,0 \\ 2x + 4y - 2z &= 6 \\ 3x + 6y - 3z &= \,9 \end{align*}

The solution was:

infinitely many solutions (x, 3, z) where x = z − 3; y = 3; z is any real number

I've spent hours on the problem. The textbook just gave a vague explanation and I can't seem to get how it works. Can someone please intuitively explain how this is?

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  • $\begingroup$ The third equation is $\dfrac32$ times the second, so isn't independent; it's like two equations, three unknowns $\endgroup$ – J. W. Tanner Jul 15 at 11:05
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    $\begingroup$ Each equation is a plane, but the plane of the second equation is the same as the plane of the third equation. So the solution of the system is the intersection of two planes: a line, which has infinitely many points. $\endgroup$ – ajotatxe Jul 15 at 11:07
  • $\begingroup$ @ajotatxe but what does "where x = z - 3; y = 3; z is any real number" mean? $\endgroup$ – Joe Jul 15 at 11:10
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Hint: Dividing the second equation by $2$ and the third by $3$ we get $$x+y-z=0$$ $$x+2y-z=3$$ $$x+2y-z=3$$ the second and the third equation are the same. Multiplying the first equation by $-1$ and adding to the second we get $y=3$. Plgging this into the first and second equation we get $$x-z=-3$$ $$x-z=-3$$ so we obtain the solutions $$x,3,x+3$$

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  • $\begingroup$ But what does the part "where x = z - 3; y = 3; z is any real number" mean? How come it's included in the solution? $\endgroup$ – Joe Jul 15 at 11:10
  • $\begingroup$ I have added some lines for you! $\endgroup$ – Dr. Sonnhard Graubner Jul 15 at 11:17
  • $\begingroup$ But then couldn't we also say that the solution is (z - 3, 3, x + 3)? $\endgroup$ – Joe Jul 15 at 12:09
  • $\begingroup$ The solution set must contain only one parameter. $\endgroup$ – Dr. Sonnhard Graubner Jul 15 at 12:11
  • $\begingroup$ I'm sorry if I'm asking such basic questions but what does that mean? Is that a rule? $\endgroup$ – Joe Jul 15 at 12:15
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If $x,y,x\in\mathbb R$ are such that $2x+4y-2z=6$, then $\frac32(2x+4y-2z)=\frac32\times6$; in other words, $3x+6y-3z=9$, which is the thir equation. Therefore, solving that system is the same thing as solving the system$$\left\{\begin{array}{l}x+y-z=0\\2x+4y-2z=6,\end{array}\right.$$which is equivalent to$$\left\{\begin{array}{l}x+y=z\\2x+4y=2z+6.\end{array}\right.$$Solving this system, you will get the solutions that you mentioned.

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  • $\begingroup$ But what does the part "where x = z - 3; y = 3; z is any real number" mean? How come it's included in the solution? $\endgroup$ – Joe Jul 15 at 11:15
  • $\begingroup$ As I wrote, if you solve my last system, you will get that $x=z-3$ and that $y=3$. Therefore, the set of all solutions is the set of all triples $(z-3,3,z)$, with $z\in\mathbb R$. $\endgroup$ – José Carlos Santos Jul 15 at 11:18

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