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I have to solve this set and should find for $x$, $x=-z$ and for $y$, $y=z$, but I don't know how did they find those values for $x$ and $y$.

Here is the system : \begin{cases} { -5x+2y-7z = 0 } \\ { -3x-4y+z = 0 } \end{cases}

Solutions : $x=−z, y=z, z\in\mathbb{R}$.

SacAndSac
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minus101
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  • What were your own attempts ? – Kurt G. Apr 08 '22 at 20:40
  • usually I know how to solve sets of equations like these but I couldn't do it with this one. Could you please help me out? – minus101 Apr 08 '22 at 20:42
  • This is a pretty straightforward problem in solving systems of equations. I would recommend the Elimination method. Start with eliminating the y variable – Vaekor Apr 08 '22 at 20:45

2 Answers2

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Remark that the two equations are independent so the dimension of the set of the solutions is $3-2=1$.

Now, let's call $(1)$ your first equation and $(2)$ your second equation. Try

  • $(1)-\frac{5}{3}(2)$.
  • $2(1)+(2)$.

Can you continue from here ?

SacAndSac
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  1. Set the two equations equal to each other.

    -5x + 2y - 7z = -3x - 4y + z

  2. Combine like terms and get x alone

    2x = 6y - 8z

    x = 3y - 4z

  3. Plug in for x in one of the equations and get y alone (I chose -3x - 4y + z = 0)

    -3(3y - 4z) - 4y + z = 0

    -9 y + 12z - 4y + z = 0

    3y = 3z

    y = z

  4. Plug in for y in x = 3y - 4z

    x = 3(z) - 4z

    x = -z

  5. Done!!!!

Vaughn von Vaught
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