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Solve the system using elimination

$-6x -2y + 3z = 34$

$-5x -4y + 4z = 32$

$2x +5y -4z = -19$

$x = ?, y = ? , z = ?$

So I threw this in a augmented matrix and put it in REF form using the row operations of:

$1) R_1 =2R_1 - R_2$

$2) R_3 = R_2 - R_3$

$3) R_3 = R_1 - R_3$

$4) R_2 = -5/7R_1 + R_2$

$5) R_3 = 9/4 R_2 + R_3$

$6) R_3 = 14/109 R_3$

To get the matrix in REF:

$\begin{bmatrix} -7&0&2&36\\0&-4&18/7&44/7\\0&0&1&12/109\end{bmatrix}$

Anyone see my mistake? How is $z = 12/109$ not a solution?

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  • $\begingroup$ I find the order in which you perform the row operations hard to follow. From the moment you started with the fraction 5/7, I quit. No need to use fractions in Gauss elimination. But if you multiply the first row by 5 and the second row by 6 and then for the new second row you would do R2 - R1, then that creates a zero under the 7 immediately. Now if you multiply row 3 by 3 and add the first row for the new third row, you get the next zero... $\endgroup$ – imranfat Jun 23 '16 at 18:28
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(Using elimination): $$-6x-2y+3z=34,~-5x-4y+4z=32,~2x+5y-4z=-19$$ $$-30x-10y+15z=170,~-30x-24y+24z=192,~2x+5y-4z=-19$$ $$-30x-10y+15z=170,~-14y+9z=22,~65y-45z=-115$$ $$-30x-10y+15z=170,~-910y+585z=1430,~-45z=180$$ $$-45z=-180\Rightarrow z=4$$ $$-910y+585\cdot4=1430\Rightarrow y=1$$ $$-30x-10\cdot1+15\cdot4=170\Rightarrow x=-4$$

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  • $\begingroup$ How do you get $65y$???? $\endgroup$ – Shammy Jun 23 '16 at 19:23
  • $\begingroup$ What is going on at line 3 , the last part $65y - 45z = -115$ makes no sense $\endgroup$ – Shammy Jun 23 '16 at 19:25
  • $\begingroup$ $$-30x-24y+24z=192$$ $$-$$ $$\underline{-30x-10y+15z=170}$$ $$-14y+9z=22$$ So: $$-30x-10y+15z=170,~-14y+9z=22,~30x+75y-60z=-285$$ Then: $$30x+75y-60z=-285$$ $$+$$ $$\underline{-30x-10y+15z=170}$$ $$65y-45z=-115$$ $\endgroup$ – Colbi Jun 23 '16 at 19:34
  • $\begingroup$ I just don't get how you guys figure out what to do so fast $\endgroup$ – Shammy Jun 23 '16 at 19:49
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consider the system $$2x+5y-4z=-19$$ $$-5x-4y+4z=32$$ $$-6x-2y+3z=34$$ multiplying the first equation by $5$ and the second by $2$ we get $$17y-12z=-31$$ (I) and the first by $3$ and adding the last we obtain $$13y-9z=-23$$ (II) $-3\cdot (I)$ plus $4\cdot (II)$ we get $y=1$

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