So the first step is to make the red three in the

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Unformatted text preview: equation can be quickly solved to find that x = 2 . Once we know this we can plug this into the second equation and that will give us an equation that we can solve for z as follows. 5 ( 2 ) + 5 z = 15 10 + 5 z = 15 5z = 5 z =1 Finally, we can substitute both x and z into the first equation which we can use to solve for y. Here is that work. 2 - 2 y + 3 (1) = 7 -2 y + 5 = 7 -2 y = 2 y = -1 So, the solution to this system is x = 2 , y = -1 and z = 1 . That was a fair amount of work and in this case there was even less work than normal since in each case we only had to multiply a single equation to allow us to eliminate variables. On top of that none of the solutions were fractions. The third method for solving systems that we’ll be looking at in the next section is really just a shorthand for what we did here, but it will be easier to do once you get used to the notation. Interpretation of solutions in these cases is a little harder in some senses. All three of these equations are the equations of planes i...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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