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by: Juan Ricker

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Linear Systems with Three Variables MATH-UA 121

Marketplace > New York University > MATH-UA 121 > Linear Systems with Three Variables
Juan Ricker
NYU

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Linear Systems with Three Variables
COURSE
Calculus I
PROF.
Vindya Vasanth Bhat
TYPE
Class Notes
PAGES
6
WORDS
KARMA
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This 6 page Class Notes was uploaded by Juan Ricker on Wednesday June 15, 2016. The Class Notes belongs to MATH-UA 121 at New York University taught by Vindya Vasanth Bhat in Summer 2016. Since its upload, it has received 14 views.

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Date Created: 06/15/16
Example 1  Solve the following system of equations.                                                          Solution We are going to try and find values of x, y, and a z that will satisfy all three equations at the  same time.  We are going to use elimination to eliminate one of the variables from one of the  equations and two of the variables from another of the equations.  The reason for doing this  will be apparent once we’ve actually done it.   The elimination method in this case will work a little differently than with two equations.  As  with two equations we will multiply as many equations as we need to so that if we start adding pairs of equations we can eliminate one of the variables.   In this case it looks like if we multiply the second equation by 2 it will be fairly simple to  eliminate the y term from the second and third equation by adding the first equation to both of  them.  So, let’s first multiply the second equation by two.                          Now, with this new system we will replace the second equation with the sum of the first and  second equations and we will replace the third equation with the sum of the first and third  equations.   Here is the resulting system of equations.                                                               So, we’ve eliminated one of the variables from two of the equations.  We now need to  eliminate either x or z from either the second or third equations.  Again, we will use  elimination to do this.  In this case we will multiply the third equation by ­5 since this will  allow us to eliminate zfrom this equation by adding the second onto is.                           Now, replace the third equation with the sum of the second and third equation.                                                                Now, at this point notice that the third equation can be quickly solved to find that  .  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.                                                                   Finally, we can substitute both x and z into the first equation which we can use to solve for y.   Here is that work.                                                              So, the solution to this system is  ,   and  .   That was a fair amount of work and in this case there was even less work than normal because in  each case we only had to multiply a single equation to allow us to eliminate variables.    In the next section we’ll be looking at a third method for solving systems that is basically a  shorthand method for what we did in the previous example.  The work using that method will be  messy as well, but it will be slightly easier to do once you get the hang of it.   In the previous example all we did was use the method of elimination until we could start solving for the variables and then just back substitute known values of variables into previous equations  to find the remaining unknown variables.   Not every linear system with three equations and three variables uses the elimination method  exclusively so let’s take a look at another example where the substitution method is used, at least partially.   Example 2  Solve the following system of equations.                                                          Solution Before we get started on the solution process do not get excited about the fact that the second  equation only has two variables in it.  That is a fairly common occurrence when we have more  than two equations in the system.    In fact, we’re going to take advantage of the fact that it only has two variables and one of  them, the y, has a coefficient of ­1.  This equation is easily solved for y to get,                                                                        We can then substitute this into the first and third equation as follows,                                                           Now, if you think about it, this is just a system of two linear equations with two variables  (x and z) and we know how to solve these kinds of systems from our work in the previous  section.   First, we’ll need to do a little simplification of the system.                                     The simplified version looks just like the systems we were solving in the previous section.   Well, it’s almost the same.  The variables this time are x and z instead of x and y, but that  really isn’t a difference.  The work of solving this will be the same.   We can use either the method of substitution or the method of elimination to solve this new  system of two linear equations.    If we wanted to use the method of substitution we could easily solve the second equation  for z (you do see why it would be easiest to solve the second equation for z right?) and  substitute that into the first equation.  This would allow us to find x and we could then find  both z and y.   However, to make the point that often we use both methods in solving systems of three linear  equations let’s use the method of elimination to solve the system of two equations.  We’ll just  need to multiply the first equation by 3 and the second by 5.  Doing this gives,                                        We can now easily solve for x to get  .  The coefficients on the second  equation are smaller so let’s plug this into that equation and solve for z.  Here is that work.                                                                   Finally, we need to determine the value of y.  This is very easy to do.  Recall in the first step  we used substitution and in that step we used the following equation.                                                                        Since we know the value of x all we need to do is plug that into this equation and get the value  of y.                                                                Note that in many cases where we used substitution on the very first step the equation you’ll  have at this step will contain both x’s and z’s and so you will need both values to get the third  variable.   Okay, to finish this example up here is the solution :  ,   and  .

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