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I have been doing some math to get ready for my first semester and I've tried doing some linear equations.I've ran into this system that I cant seem to solve. \begin{array}{r c r c r c r} x & - & 2y & + & 3z & = & -2\\ 4x & - & y & + & z & = & 3\\ 3x & + & 2y & - & z & = & 4 \end{array}

I've tried multiplying the first equation with $4$ and $3$ to eliminate $x$, but I'm not able to get further with that even after getting rid of the $x$, I still have 2 unknowns in $y$ and $z$. I can't seem to get rid of $y$ or $z$. I've also tried substitution (not sure if that is how you say it in English) but the same story. What am I missing here?

Thanks!

N. F. Taussig
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codeisfun
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  • try using Matrix if you are familiar with – Predator Monarch Oct 04 '20 at 14:56
  • Take two equations at a time (first two and then last two). Try and come up with two equations in two variables by eliminating one variable (the same one both times). – Math Lover Oct 04 '20 at 14:58
  • If you dont like/ are unfamiliar with matrices (which is apparent), you can find (x,y) in terms of z from any 2 of the three equations, then input (x,y) in the last untouched equation to get z, substitute z and enjoy...(CAUTION: maynot work for an inconsistent system, vigilance required) – Anindya Prithvi Oct 04 '20 at 15:11
  • I've done what Math Lover has recommended and I've fond a solution. I've added the 1st and 3rd equation by that eliminating y, and i've multipled the 2nd one with -2 and added it to the first also eliminating the y.Than I could easily eliminate z and get X, put it back in the prior equations and gotten the results. Thank you for the help! – codeisfun Oct 04 '20 at 15:16

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