 My Math Forum I have to solve 3 equations with 3 variables and I'm extremely confused
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 January 3rd, 2016, 04:03 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, 04:18 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 $2y + 6z = 0$. Thanks from nsganon101 January 3rd, 2016, 04:26 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 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, 04:32 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, 04:50 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 self-learning) January 3rd, 2016, 05:25 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 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, 05:25 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, 05:27 PM #8 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 No, one normally eliminates one variable at a time, but I noticed a way to eliminate y and z together. January 3rd, 2016, 05:32 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, 05:58 PM #10 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 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. Tags confused, equations, extremely, solve, variables Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post fixxit Algebra 4 September 10th, 2014 11:29 AM nakota2k Calculus 3 February 1st, 2014 04:18 AM alexandra2727 Algebra 5 January 16th, 2014 10:38 AM Crystal-Field Calculus 3 April 29th, 2009 06:12 PM IsenseHelp Algebra 2 January 11th, 2009 04:10 PM

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