January 3rd, 2016, 05:03 PM  #1 
Newbie Joined: Jan 2016 From: Oregon Posts: 5 Thanks: 0  I have to solve 3 equations with 3 variables and I'm extremely confused
the 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. 
January 3rd, 2016, 05:18 PM  #2 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
That means $x$ multiplied by 0. Essentially you can ignore it and leave the equation as $2y + 6z = 0$.

January 3rd, 2016, 05:26 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,950 Thanks: 1842 
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. 
January 3rd, 2016, 05:32 PM  #4 
Newbie Joined: Jan 2016 From: Oregon Posts: 5 Thanks: 0 
what do you mean "from the first equation?" Aren't I supposed to add two equations together?

January 3rd, 2016, 05:50 PM  #5 
Newbie Joined: Jan 2016 From: Oregon Posts: 5 Thanks: 0 
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 selflearning) 
January 3rd, 2016, 06:25 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,950 Thanks: 1842 
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. 
January 3rd, 2016, 06:25 PM  #7 
Newbie Joined: Jan 2016 From: Oregon Posts: 5 Thanks: 0 
so whenever I do this I should always double the first equation that I try to combine?

January 3rd, 2016, 06:27 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 19,950 Thanks: 1842 
No, one normally eliminates one variable at a time, but I noticed a way to eliminate y and z together.

January 3rd, 2016, 06:32 PM  #9 
Newbie Joined: Jan 2016 From: Oregon Posts: 5 Thanks: 0 
(sorry if I'm being a pest) How did you get the idea to subtract 2x  7y  z = 8 from the first equation? 
January 3rd, 2016, 06:58 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 19,950 Thanks: 1842 
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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confused, equations, extremely, solve, variables 
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