we subtract the first equation from the third to eliminate y
![\begin{gathered} x+2y+0z=-1 \\ -4x+2y-3z=-30 \\ ---------------- \\ 5x+0y+3z=29 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fyj35gkhf70j64i0s0hsvoqy4shp59t7we.png)
we multiply the second equation with 2 and sum it to the third, to eliminate y too
![\begin{gathered} 6x-2y+8z=34 \\ -4x+2y-3z=-30 \\ ---------------- \\ 2x+0y+5z=4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ppunpb58ydnddav01qq0tpj9m35ikty1v7.png)
the new equations are the fourth and fifth respectively
Now we will eliminate Z from these two equations
we multiply the fourth equation with 5/3
![\begin{gathered} 5x((5)/(3))+3z((5)/(3))=29((5)/(3)) \\ (25)/(3)x+5z=(145)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7g0ughmgjomxj8h841vlfe1snq8zjn98om.png)
We can name this the sixth equation but it is the same what the fourth
we subtract the sixth equation from the fifth to eliminate z
![\begin{gathered} (25)/(3)x+5z=(145)/(3) \\ 2x+5z=4 \\ ------------ \\ (19)/(3)x+0z=(133)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/sbj4g08d26hagcor69kdwk45tqq86vnnp4.png)
now we can solve X and replace in the other equations to find Y and Z
![\begin{gathered} (19)/(3)x=(133)/(3) \\ x=(133*3)/(19*3) \\ x=7 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8la9j5olokkp37occq72a8xqvk3pjt9q35.png)
replace x=7 on the fiste equation to find Y
![\begin{gathered} (7)+2y=-1 \\ 2y=-1-7 \\ y=(-8)/(2) \\ y=-4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hltruiaak0yndqyl8nirabgpglo6gwa01k.png)
replace y=-4 and x=7 on the second equation to find Z
![\begin{gathered} 3(7)-(-4)+4z=17 \\ 21+4+4z=17 \\ 4z=17-21-4 \\ z=(-8)/(4) \\ z=-2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/sjbsom7btbzf8l1shu1g8jxtcg9fv7e8pt.png)
you can check replacing the values in any equation and the equality must be satisfied