- **Algebra**
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- - **I have to solve 3 equations with 3 variables and I'm extremely confused**
(*http://mymathforum.com/algebra/311677-i-have-solve-3-equations-3-variables-im-extremely-confused.html*)

I have to solve 3 equations with 3 variables and I'm extremely confusedthe three equations are: 4x + 3y -z = 4 2x - 9y - 7z = 8 (0x) + 2y + 6z = 0 I have gotten to the point of understanding that I need to combine two of the equations and to also eliminate one variable. One of the biggest issues is I don't understand "(0x)," I've never seen that before. I feel like once I understand what that is I will be able to solve this problem. |

That means $x$ multiplied by 0. Essentially you can ignore it and leave the equation as $2y + 6z = 0$. |

From the first equation, 8x + 6y - 2z = 8. Adding the second equation gives 10x - 3y - 9z = 16. From the third equation, -3y - 9z = 0, so 10x = 16. |

what do you mean "from the first equation?" Aren't I supposed to add two equations together? |

so I tried adding the first two equations and got: 6x - 6y - 6z = 12 adding the last two I get: 2x - 7y - z = 8 when I try to combine the two I get: 8x + y - 5z = 20 I don't know what to do with that because I thought I was supposed to get rid of one of the variables... (sorry I'm self-learning) |

I doubled the first equation, then added the second equation to the result. That enabled me to use the third equation to eliminate the variables y and z together, leaving 10x = 16, i.e., x = 1.6. After you obtained 2x - 7y - z = 8, you could subtract the first equation to get -2x - 10y = 4, which implies y = (-2x - 4)/10 = (-3.2 - 4)/10 = -0/72. You made a slip when you added the first two equations. You should have got 6x - 6y - 8z = 12. |

so whenever I do this I should always double the first equation that I try to combine? |

No, one normally eliminates one variable at a time, but I noticed a way to eliminate y and z together. |

(sorry if I'm being a pest) How did you get the idea to subtract 2x - 7y - z = 8 from the first equation? |

I did that to eliminate z. There are other ways to eliminate z. Many solvers would have solved by using the following method, which I give for comparison purposes. From the third equation, y = -3z. Substituting -3z for y in the first two equations gives 4x - 9z - z = 4 and 2x + 27z - 7z = 8, so 4x - 10z = 4 and 2x + 20z = 8. Doubling the first of those equations and then adding the second gives 10x = 16, as obtained before, allowing one to say 4(1.6) - 10z = 4, so that z = (4(1.6) - 4)/10 = 0.24, and then y = -3z = -0.72. Alternatively, doubling the second and then subtracting the first would give 50z = 12, so z = 0.24. Again, y = -3z = -0.72, and x can be calculated as (4 + 10z)/4 = 6.4/4 = 1.6. |

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