8.7 Systems of Non-Linear Equations and Inequalities

Transcription

1 8.7 Sstems of Non-Linear Equations and Inequalities Sstems of Non-Linear Equations and Inequalities In this section, we stud sstems of non-linear equations and inequalities. Unlike the sstems of linear equations for which we have developed several algorithmic solution techniques, there is no general algorithm to solve sstems of non-linear equations. Moreover, all of the usual hazards of non-linear equations like etraneous solutions and unusual function domains are once again present. Along with the tried and true techniques of substitution and elimination, we shall often need equal parts tenacit and ingenuit to see a problem through to the end. You ma find it necessar to review topics throughout the tet which pertain to solving equations involving the various functions we have studied thus far. To get the section rolling we begin with a fairl routine eample. Eample Solve the following sstems of equations. Verif our answers algebraicall and graphicall.. { + = + 9 = 6 { + = 9 = 6. { + = = 0 { + = = 0.. Solution:. Since both equations contain and onl, we can eliminate one of the variables as we did in Section 8.. { E) + = E) + 9 = 6 Replace E with E + E { E) + = E) = 0 From = 0, we get = or = ±. To find the associated values, we substitute each value of into one of the equations to find the resulting value of. Choosing + =, we find that for both = and =, we get = 0. Our solution is thus {0, ), 0, )}. To check this algebraicall, we need to show that both points satisf both of the original equations. We leave it to the reader to verif this. To check our answer graphicall, we sketch both equations and look for their points of intersection. The graph of + = is a circle centered at 0, 0) with a radius of, whereas the graph of +9 = 6, when written in the standard form 9 + = is easil recognized as an ellipse centered at 0, 0) with a major ais along the -ais of length 6 and a minor ais along the -ais of length. We see from the graph that the two curves intersect at their -intercepts onl, 0, ±).. We proceed as before to eliminate one of the variables { E) + = E) 9 = 6 Replace E with E + E { E) + = E) = 0

2 68 Sstems of Equations and Matrices Since the equation = 0 admits no real solution, the sstem is inconsistent. To verif this graphicall, we note that + = is the same circle as before, but when writing the second equation in standard form, 9 =, we find a hperbola centered at 0, 0) opening to the left and right with a transverse ais of length 6 and a conjugate ais of length. We see that the circle and the hperbola have no points in common. Graphs for { + = + 9 = 6 Graphs for { + = 9 = 6. Since there are no like terms among the two equations, elimination won t do us an good. We turn to substitution and from the equation = 0, we get =. Substituting this into + = gives + ) =. Solving, we find = or = ±. Returning to the equation we used for the substitution, =, we find = when = ) ), so one solution is,. Similarl, we find the other solution to be,. We leave it to the reader that both points satisf both equations, so that our final answer is {, ),, )}. The graph of + = is our circle from before and the graph of = 0 is a line through the origin with slope. Though we cannot verif the numerical values of the points of intersection from our sketch, we do see that we have two solutions: one in Quadrant I and one in Quadrant III as required.. While it ma be tempting to solve = 0 as = and substitute, we note that this sstem is set up for elimination. { E) + = E) = 0 Replace E with E + E { E) + = E) + = From + = we get + = 0 which gives = ± 7. Due to the complicated nature of these answers, it is worth our time to make a quick sketch of both equations to head off an etraneous solutions we ma encounter. We see that the circle + = intersects the parabola = eactl twice, and both of these points have a positive value. Of the two solutions for, onl = + 7 is positive, so to get our solution, we substitute this We encourage the reader to solve the sstem using substitution to see that ou get the same solution.

3 8.7 Sstems of Non-Linear Equations and Inequalities 69 into + 7 = 0 and solve for. We get = ± { + 7, + 7 ), + 7, Our solution is = ± )}, which we leave to the reader to verif. Graphs for { + = = 0 Graphs for { + = = 6 A couple of remarks about Eample 8.7. are in order. First note that, unlike sstems of linear equations, it is possible for a sstem of non-linear equations to have more than one solution without having infinitel man solutions. In fact, while we characterize sstems of nonlinear equations as being consistent or inconsistent, we generall don t use the labels dependent or independent. Secondl, as we saw with number, sometimes making a quick sketch of the problem situation can save a lot of time and effort. While in general the curves in a sstem of non-linear equations ma not be easil visualized, it sometimes pas to take advantage when the are. Our net eample provides some considerable review of man of the topics introduced in this tet. Eample Solve the following sstems of equations. Verif our answers algebraicall and graphicall, as appropriate. {. + 6 = = 0 Solution. { + e = + e =. z ) = z = ) + =. At first glance, it doesn t appear as though elimination will do us an good since it s clear that we cannot completel eliminate one of the variables. The alternative, solving one of the equations for one variable and substituting it into the other, is full of unpleasantness. Returning to elimination, we note that it is possible to eliminate the troublesome term, and the constant term as well, b elimination and doing so we get a more tractable relationship between and { E) + 6 = 0 E) + 6 = 0 Replace E with E + E { E) + 6 = 0 E) = 0

4 60 Sstems of Equations and Matrices We get = 0 or = ±. Substituting = into E we get + 6 = 0 so that = 6 or = ±. On the other hand, when we substitute = into E, we get 6 = 0 or = 6 which gives no real solutions. Substituting each of = ± { ) )} into the substitution equation = ields the solution,,,. We leave it to the reader to show that both points satisf both equations and now turn to verifing our solution graphicall. We begin b solving + 6 = 0 for to obtain = 6. This function is easil graphed using the techniques of Section.. Solving the second equation, + 6 = 0, for, however, is more complicated. We use the quadratic formula to obtain = ± + 6 which would require the use of Calculus or a calculator to graph. Believe it or not, we don t need either because the equation + 6 = 0 can be obtained from the equation + 6 = 0 b interchanging and. Thinking back to Section., this means we can obtain the graph of + 6 = 0 b reflecting the graph of + 6 = 0 across the line =. Doing so confirms that the two graphs intersect twice: once in Quadrant I, and once in Quadrant III as required. The graphs of + 6 = 0 and + 6 = 0. Unlike the previous problem, there seems to be no avoiding substitution and a bit of algebraic unpleasantness. Solving + e = for, we get = e which, when substituted into the second equation, ields e ) + e =. After epanding and gathering like terms, we get 6e 8e + e = 0. Factoring gives us e 8e e + ) = 0, and since e 0 for an real, we are left with solving 8e e + = 0. We have three terms, and even though this is not a quadratic in disguise, we can benefit from the substitution u = e. The equation becomes 8u u+ = 0. Using the techniques set forth in Section., we find u = is a zero and use snthetic division to factor the left hand side as u ) 8u + u ). We use the quadratic formula to solve 8u + u = 0 and find u = ±. Since u = e, we now must solve e = and e = ±. From e =, we get = ln ) = ln). As for e = ±, we first note that < 0, so e = has no real solutions. We are

5 8.7 Sstems of Non-Linear Equations and Inequalities 6 left with e = + ), so that = ln +. We now return to = e to find the accompaning values for each of our solutions for. For = ln), we get For = ln ) +, we have = e = e ln) = e ln ) = ) = 0 We get two solutions, { 0, ln)), ln = e = e ln + = e ln + = = = + + ) ) ) + ) 8 ) )}, +. It is a good review of the properties of logarithms to verif both solutions, so we leave that to the reader. We are able to sketch = e using transformations, but the second equation is more difficult and we resort to the calculator. We note that to graph + e =, we need to graph both the positive and negative roots, = ± e. After some careful zooming, we get The graphs of = e and = ± e.. Our last sstem involves three variables and gives some insight on how to keep such sstems organized. Labeling the equations as before, we have The calculator has trouble confirming the solution ln), 0) due to its issues in graphing square root functions. If we mentall connect the two branches of the thicker curve, we see the intersection.

6 6 Sstems of Equations and Matrices E z ) = E z = E ) + = The easiest equation to start with appears to be E. While it ma be tempting to divide both sides of E b, we caution against this practice because it presupposes 0. Instead, we take E and rewrite it as z = 0 so z ) = 0. From this, we get two cases: = 0 or z =. We take each case in turn. Case : = 0. Substituting = 0 into E and E, we get { E z ) = E ) = Solving E for gives = or =. Substituting these values into E gives z = when = and z = when =. We obtain two solutions,, 0, ) and, 0, ). Case : z =. Substituting z = into E and E gives us { E ) ) = E ) + = Equation E gives us = or = 0, which is a contradiction. This means we have no solution to the sstem in this case, even though E is solvable and gives = 0. Hence, our final answer is {, 0, ),, 0, )}. These points are eas enough to check algebraicall in our three original equations, so that is left to the reader. As for verifing these solutions graphicall, the require plotting surfaces in three dimensions and looking for intersection points. While this is beond the scope of this book, we provide a snapshot of the graphs of our three equations near one of the solution points,, 0, ). Eample 8.7. showcases some of the ingenuit and tenacit mentioned at the beginning of the section. Sometimes ou just have to look at a sstem the right wa to find the most efficient method to solve it. Sometimes ou just have to tr something.

7 8.7 Sstems of Non-Linear Equations and Inequalities 6 We close this section discussing how non-linear inequalities can be used to describe regions in the plane which we first introduced in Section.. Before we embark on some eamples, a little motivation is in order. Suppose we wish to solve <. If we mimic the algorithms for solving nonlinear inequalities in one variable, we would gather all of the terms on one side and leave a 0 on the other to obtain + < 0. Then we would find the zeros of the left hand side, that is, where is + = 0, or + =. Instead of obtaining a few numbers which divide the real number line into intervals, we get an equation of a curve, in this case, a circle, which divides the plane into two regions - the inside and outside of the circle - with the circle itself as the boundar between the two. Just like we used test values to determine whether or not an interval belongs to the solution of the inequalit, we use test points in the each of the regions to see which of these belong to our solution set. We choose 0, 0) to represent the region inside the circle and 0, ) to represent the points outside of the circle. When we substitute 0, 0) into + < 0, we get < which is true. This means 0, 0) and all the other points inside the circle are part of the solution. On the other hand, when we substitute 0, ) into the same inequalit, we get < 0 which is false. This means 0, ) along with all other points outside the circle are not part of the solution. What about points on the circle itself? Choosing a point on the circle, sa 0, ), we get 0 < 0, which means the circle itself does not satisf the inequalit. As a result, we leave the circle dashed in the final diagram. The solution to < We put this technique to good use in the following eample. Eample Sketch the solution to the following nonlinear inequalities in the plane. {. < Solution.. The inequalit < + is a compound inequalit. It translates as and < +. As usual, we solve each inequalit and take the set theoretic intersection to determine the region which satisfies both inequalities. To solve, we write The theor behind wh all this works is, surprisingl, the same theor which guarantees that sign diagrams work the wa the do - continuit and the Intermediate Value Theorem - but in this case, applied to functions of more than one variable. Another wa to see this is that points on the circle satisf + = 0, so the do not satisf + < 0.

8 6 Sstems of Equations and Matrices 0. The curve = 0 describes a parabola since eactl one of the variables is squared. Rewriting this in standard form, we get = + and we see that the verte is, 0) and the parabola opens to the right. Using the test points, 0) and 0, 0), we find that the solution to the inequalit includes the region to the right of, or inside, the parabola. The points on the parabola itself are also part of the solution, since the verte, 0) satisfies the inequalit. We now turn our attention to < +. Proceeding as before, we write < 0 and focus our attention on = 0, which is the line =. Using the test points 0, 0) and 0, ), we find points in the region above the line = satisf the inequalit. The points on the line = do not satisf the inequalit, since the -intercept 0, ) does not. We see that these two regions do overlap, and to make the graph more precise, we seek the intersection of these two curves. That is, we need to solve the sstem of nonlinear equations { E) = + E) = Solving E for, we get =. Substituting this into E gives =, or 6 = 0. We find = and = and since =, we get that the graphs intersect at 0, ) and, ). Putting all of this together, we get our final answer below. < + < +. To solve this sstem of inequalities, we need to find all of the points, ) which satisf both inequalities. To do this, we solve each inequalit separatel and take the set theoretic intersection of the solution sets. We begin with the inequalit + which we rewrite as + 0. The points which satisf + = 0 form our friendl circle + =. Using test points 0, 0) and 0, ) we find that our solution comprises the region outside the circle. As far as the circle itself, the point 0, ) satisfies the inequalit, so the circle itself is part of the solution set. Moving to the inequalit + 0, we start with + = 0. Completing the squares, we obtain ) + ) =, which is a circle centered at, ) with a radius of. Choosing, ) to represent the inside of the circle,, ) as a point outside of the circle and 0, 0) as a point on the circle, we find that the solution to the inequalit is the inside of the circle, including the circle itself. Our final answer, then, consists of the points on or outside of the circle + = which lie on or

9 8.7 Sstems of Non-Linear Equations and Inequalities 6 inside the circle ) + ) =. To produce the most accurate graph, we need to find where these circles intersect. To that end, we solve the sstem { E) + = E) + = 0 We can eliminate both the and b replacing E with E + E. Doing so produces =. Solving this for, we get =. Substituting this into E gives + ) = which simplifies to + + = or = 0. Factoring ields ) which gives = 0 or =. Substituting these values into = gives the points 0, ) and, 0). The intermediate graphs and final solution are below Solution to the sstem.