The Hornet Posted July 25, 2009 Share Posted July 25, 2009 Yes, this is correct. It can easily be solved using linear algebra: So, x=4 and y=11. Link to comment Share on other sites More sharing options...
DYU3ster Posted July 25, 2009 Share Posted July 25, 2009 Good to know I am still able to do Math 11 lol Link to comment Share on other sites More sharing options...
Scrub Posted July 25, 2009 Share Posted July 25, 2009 Yes, this is correct. It can easily be solved using linear algebra: So, x=4 and y=11. Link to comment Share on other sites More sharing options...
Sahota9 Posted July 25, 2009 Share Posted July 25, 2009 thanks guys Link to comment Share on other sites More sharing options...
Sahota9 Posted July 25, 2009 Share Posted July 25, 2009 I found the lowest Common multiple and converted the equations to: 15x + 2y = 62 8x - 15y = 17 So what is the next step? How am I supposed to remove on of the variables any ideas? Thanks. Link to comment Share on other sites More sharing options...
iPwn Posted July 25, 2009 Share Posted July 25, 2009 Solve By Elimation. 5(x+2) + y+8 = 6 --------- ----- 6 9 2x+1 - 5y = 7 ------ ---- --- 3 4 4 I found the lowest Common multiple and converted the equations to: 15x + 2y = 62 8x - 15y = 17 So what is the next step? How am I supposed to remove on of the variables any ideas? Thanks. Link to comment Share on other sites More sharing options...
Sahota9 Posted July 25, 2009 Share Posted July 25, 2009 fixed it Link to comment Share on other sites More sharing options...
Sahota9 Posted July 25, 2009 Share Posted July 25, 2009 Use the same method you were taught above Link to comment Share on other sites More sharing options...
Sahota9 Posted July 26, 2009 Share Posted July 26, 2009 bump Link to comment Share on other sites More sharing options...
Dorkey Posted July 26, 2009 Share Posted July 26, 2009 You gotta multiply one equation by a certain number (and the same with the second equation, if needed), then add to remove a variable. Link to comment Share on other sites More sharing options...
Sahota9 Posted July 26, 2009 Share Posted July 26, 2009 I figured it out thanks Link to comment Share on other sites More sharing options...
Sahota9 Posted July 27, 2009 Share Posted July 27, 2009 Can anyone help me with this 3 by 3 system by solving? x+y+2z=0 1/4x-1/4y+1/8z=1 3/2x+y+1/2z=1 I reduced them and got the equations to: 1) x+y+2z=0 2) 4x-4y+2z=16 3) 3x+2y+2z=2 Whats the next step? Link to comment Share on other sites More sharing options...
Scrub Posted July 27, 2009 Share Posted July 27, 2009 Can anyone help me with this 3 by 3 system by solving? x+y+2z=0 1/4x-1/4y+1/8z=1 3/2x+y+1/2z=1 I reduced them and got the equations to: 1) x+y+2z=0 2) 4x-4y+2z=16 3) 3x+2y+2z=2 Whats the next step? Link to comment Share on other sites More sharing options...
Sahota9 Posted July 28, 2009 Share Posted July 28, 2009 damn i tried doing it and got (-4,8,-2) and the real answer is (2,-2,0) what am i doing wrong Link to comment Share on other sites More sharing options...
Scrub Posted July 28, 2009 Share Posted July 28, 2009 damn i tried doing it and got (-4,8,-2) and the real answer is (2,-2,0) what am i doing wrong Link to comment Share on other sites More sharing options...
trek Posted July 28, 2009 Share Posted July 28, 2009 Can anyone help me with this 3 by 3 system by solving? x+y+2z=0 1/4x-1/4y+1/8z=1 3/2x+y+1/2z=1 I reduced them and got the equations to: 1) x+y+2z=0 2) 4x-4y+2z=16 3) 3x+2y+2z=2 Whats the next step? Link to comment Share on other sites More sharing options...
believe_in_blue Posted July 28, 2009 Share Posted July 28, 2009 Since you have 3 equations and 3 variables the system is solvable.. so the longest but foolproof way is to substitute one equation in to another one, and substitute again to eliminate the variables until you only have one. Once you solve one variable, substitute back in your second equation to get the second variable. Once you have the first and second variable, solving the last variable should be a piece of cake. Alternatively, if you're familiar with linear algebra, you can use elementary row operations to reduce the matrix into row echelon form which will give you the solution to the system. Example 3 of http://tutorial.math.lamar.edu/Classes/Lin...temsOfEqns.aspx is pretty much what you're looking for Link to comment Share on other sites More sharing options...
Scrub Posted July 29, 2009 Share Posted July 29, 2009 having three equations with three variables does not guarantee that the system has a solution. what about x + y + z = 1 x + y + z = 2 x + y + z = 3 (three parallel planes in 3-space which do not intersect)? Link to comment Share on other sites More sharing options...
Sahota9 Posted July 30, 2009 Share Posted July 30, 2009 4ax + 5y = 10 2x-3y=9 can you help me find a? Link to comment Share on other sites More sharing options...
Sahota9 Posted July 31, 2009 Share Posted July 31, 2009 bump Link to comment Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.