Algebra/Trig Review - Pauls Online Math Notes - Lamar University

Algebra/Trig ReviewFrom this we see that y = − 1. Plugging this into either of the above two equationsyields z = 2 . Finally, plugging both of these answers into x=−z−3y− 1 yieldsx = 0 .Method 2In the second method we add multiples of two equations together in such a way toeliminate one of the variables. We’ll do it using two different sets of equationseliminating the same variable in both. This will give a system of two equations intwo unknowns which we can solve.So we’ll start by noticing that if we multiply the second equation by -2 and add it tothe first equation we get.2x− y− 2z=−3−2x−6y− 2z= 2−7y− 4z=−1Next multiply the second equation by -5 and add it to the third equation. This gives−5x−15y− 5z= 55x− 4y+ 3z= 10−19y− 2z= 15This gives the following system of two equations.−7y− 4z=−1−19y− 2z= 15We can now solve this by multiplying the second by -2 and adding−7y− 4z=−138y+ 4z=−3031y= −31From this we get that y = − 1, the same as the first solution method. Plug this intoeither of the two equations involving only y and z and we’ll get that z = 2 . Finallyplug these into any of the original three equations and we’ll get x = 0 .You can use either of the two solution methods. In this case both methods involvedthe same basic level of work. In other cases on may be significantly easier than theother. You’ll need to evaluate each system as you get it to determine which methodwill work the best.InterpretationRecall that one interpretation of the solution to a system of equations is that thesolution(s) are the location(s) where the curves or surfaces (as in this case) intersect.So, the three equations in this system are the equations of planes in 3D space (you’lllearn this in Calculus II if you don’t already know this). So, from our solution weknow that the three planes will intersect at the point (0,-1,2). Below is a graph of thethree planes.© 2006 Paul Dawkins 33http://tutorial.math.lamar.edu/terms.aspx