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(-2)(2x - 3y) = (-2)(-7) em to sausood oldneon ay # 2 T novolog 61 3 3 Buloa numbsup "enormups issalmon to moreve eds not gnivlog m abordom and sit inw a the Tenpileups momfron to amate ?? bid to mobbloe shi meziger adding androb woll d's Systems of Nonlinear Equations vious examine Definition: A system of equations is nonlinear if it involves at least one equation that is not linear. A solution to a system of equations of two variables is an ordered pair that makes all the equations true. The collection of all solutions enclosed in braces and separated by commas is what we call a solution set. Isoidqing bas notionmile nonbluedue boeu ebordism ono m art A. Solving Systems of Equations Using Substitution redue ord 2) I o I olqmax We begin our extension with a system involving one linear equation and one quadratic equation. In this case, it is always possible to use substitution by solving the linear equation for one of the variables. Example 1.6.3. Solve the following system, and sketch the graphs in one Cartesian plane nousups T-y+2=0 y-1=x2 Solution. We solve for y in terms of r in the first equation, and substitute this expression to the second equation. -y+2=0 = y=1+2 k l+ v5 420 5+ v5 x-1-1= 0 1 lo iniog 2 mivoda sung 21ginD) san 2 dgam 1 - V5 1 - V5 5 - V5 1 + v5 I = y= - -+2= 2 2 2 Solutions: 1 + V5 5 + V5 1 - V5 5-V5 and The first equation represents a line with r-intercept -2 and y-intercept 2, while the second equation represents a parabola with vertex at (0, 1) and which opens upward.INTRODUCTION When you were in junior high school, you solved for the solution's of systems of linger equations. A short recall about this lesson is given at the beginning of the content discussion of this module to refresh your mind about the methods used to find the solutions. Then we extend these methods to solve a system of equations which are not necessarily linear. In this lesson, the equations are restricted to linear and quadratic types, although it is possible to adapt the methodology so systems with other types of equations. We focus on quadratic equations for two reasons: to include a graphical representation of the solution and to ensure that either a solution is obtained or it is determined that there is no solution. The latter is possible because of the quadratic formula. After the discussion, you should be able to answer the following questions correctly. a. What are the methods in solving for the system of nonlinear equations? b. How do the graphs represent the solutions of the systems of nonlinear equations? CONTENT DISCUSSIONS lon a ford nonsops ard Review of Techniques in Solving Systems of Linear Equations mole s of noitelee A war ow terw zi Recall the methods we used to solve systems of linear equations. There were three methods used: substitution, elimination, and graphical. Example 1.6.1. Use the substitution method to solve the system, and sketch the graphs in one sumbsup ano DOCartesian plane showing the point of intersection. we muged nousupo wod! off ginivlog Arty = 6 goldsriv bob to sno rot 5x + 3y = 4 Solution. Isolate the variable y in the first equation, and then substitute into the second equation. Arty = 6 y= 6 - Ar 5r + 3(6 - 4r) = 4 -78 + 18 = 4 r = ?' y = 6 - 4(2) = -2 Example 1.6.2. Use the elimination method to solve the system, and sketch the graphs in one Cartesian plane showing the point of intersection. 2x + 7 = 3y 4x + 7y = 12 Solution. We eliminate first the variable r. Rewrite the first equation wherein only the constant term is on the right-hand side of the equation, then multiply it by -2, and then add the resulting equation to the second equation.1, 7) 6 Dewey'-- 49- 6y - 11 5 ........ 4(3 - 2)= (1-3)' XS -6 -5 -3 5 6 .C- Usually, the general form is more convenient to use in solving systems of equations. However, sometimes the solution can be simplified by writing the equations in standard form. Moreover, the standard form is best for graphing. Let us again solve the previous example in a different way. 0900 + 34 625y Solution 2. By completing the square, we can change the first equation into stan- dard form: 32 - 4x - 6y = 11 = 4(x +5) = (y-3)2. 4(2+ 5)1=(3+3)2 up. lo amolave gniviol .H boon amsteve smoe gomismod anons 4 (3 - I) = (4 - 3)2 m hitsen oals et borbor noumanimita ] Using substitution or the transitive property of equality, we get due) asupinmost rod 4(1+5)=4(3- x) =lollol sill ovlod Adri sigmaxed Example 1.6.5. Solve the system and graph the curves: (1 - 3)2 + (y - 5)2 = 10 vd s pidstay eds stagimai x2 + (y + 1)? = 25:092 ads baaqxo oW .I naimlol. anoldsupe ond gibbs Solution. Expanding both equations, we obtain - SI x' + y2 - 61 - 10y + 24 = 0 12 + y2 + 2y - 24 =0. Subtracting these two equations, we get of equations where at least one of dung boom Solved using the -6: - 12y + 48 = 0 - I + 2y -8 =0 We can substitute x = 8- 2y to either the first equation or the second equation. For convenience, we choose the second equation. -(8 -29)3+ y+2y -24-0 0Given Let y 3 2 y-1=2' 2 7 7 : 1-V5 5-V5) 2 did memore that enter a X -4 -3 -2 -1. 0 1 2 3 4 5 6 Hou8 ed hi mot bisbrad movesold mol bebasis ni mollaups you toowetlib a at slumaxe anolong add svloe nings au da.] carrot brab B. Solving Systems of Equations Using Elimination Elimination method is also useful in systems of nonlinear equations. Sometimes, some systems need both techniques ( substitution and elimination) to solve them. anu off 10 nottutredus quiet Example 1.6.4. Solve the following system: (1 E) = (8 + 1)A y' - 41 - 6y = 11 modatandi svlo2 .3.0.1 olqmixed 4(3 - 1) = (y-3)2. Solution 1. We expand the second equation, and eliminate the variable r by adding the equations. disido ow anoisups dood quibmngx! soruled 4(3 -1) = (y -3)2 = 12-4x = y' -by + 9 = y' + 4x -by = 3 0=10 + 501 -30 y2 - 4x - 6y = 11 y' + 41 -6y = 3 Adding these equations, we get 2y?- 12y = 14 => y2-6y-7=0 = (y-7)(y+1) =0 - y=7 ory = -1. Solving for a in the second equation, we have noiteups ((y e-3)? seoods ow sonsinovnoo to1 1 =3- y =7 - -1 and = = Solutions: (-1,7) and (-1,-1) The graphs of the equations in the preceding example with the points of intersection are shown below.Given Let x- width of the TV screen; y - length of the TV screen; Area = 300 square inches; 71 Required: length and width of the TV screen ( Equation: Diagonal of the TV screen is 25 inches x ty = 25' X'ty = 625 The area of the TV screen is 300 square inches x' y =300 Solution: S 90000 x + y' = 625 + y = 625 y x. y =300 90000 + y* = 625y x. y =300 solve for x y* - 625y' + 90000 = 0 _300 X =3 (y' - 225)(y' - 400) = 0 y y' - 225 = 0 y' - 400 = 0 x + y = 625 substitute x = 300 aupd maniino y'= 225 8 boy'= 400 qq y = $15 y = +20 ignol on estgor 30924 v2 2 625 novig arelop noteivalof to axis god esquede insistlib ni smoo neo svie amse andi'd 199792 Isho idglad end of shibe and to one ads yousaba e of our oxy= 300 xy = 300 LEN of outer boogze masse end VT CI: I don-Of A & Polls Josex . 15 - 300 x . 20 = 300 blu noone off to rubix = 20 u dign x - 15 1 If the length is 15 inches, the width is 20 inches. moleva sill ousd ow mood? If the length is 20 inches, the width is 15 inches. 0031 ESSENTIAL LEARNING/SUMMARY 0001 2 SA - Sea System of Nonlinear Equations - a system of equations where at least one of the equations is nonlinear. Like solving systems of linear equations, the system of nonlinear equations can be solved using three methods. a) Substitution - involves solving for one of the variables in any of the given equations and substituting the equivalent expression for the variable in the other equation. b) Elimination - involves eliminating a variable in the equations which has the same coefficients but opposite signs. c) Graphical - involves sketching the graphs of the given equations in the same rectangular coordinate system. The point(s) where the graphs intersect determine($) arm ." the solution(s) to the given system. mages afw The solution of a system of nonlinear equations is primarily in the form of an ordered pair. bonfromThe ordered pair must satisfy both of the original equations in the given system to consider it sit sin a solution. Thus, it is always important to check whether the ordered pair satisfies the too equations in the system.#=2 = =8-2(2) =4 - and y=4 = 18 24)-0 The solutions are (4, 2) and (0, 4). The graphs of both equations are circles, One has center (3, 5) and radius 10, while the other has center (0, -1) and radius 5. The graphs with the points of intersection are show below. nottoup ! (0,4) 21 (4.2) 3421-8=0 -4 -2 10 2 4 6 8 40 12 14 -2 -4 25dy + 000De 2 + (y+ 1) =25 0= 00008 + 258 . x not ovloe 008 = v x Applications of Systems of Nonlinear Equations 214 Example 1.6.6. The screen size of television sets is given in inches. This indicates the length of the diagonal. Screens of the same size can come in different shapes. Wide-screen TV's usually have screens with aspect ratio 16 : 9, indicating the ratio of the width to the height. Older TV models often have aspect ratio 4 : 3. A 40-inch LED TV has screen aspect ratio 16 : 9. Find the length and the width of the screen. Solution. Let w represent the width and h the height of the screen. Then, by Pythagorean, A Theorem, we have the system Meortoni Of a dibiw och pordont 2 1 or dignial ondr 11 eordom ? I ai thorw sch esdom Of at dranof andi'll wth = 402 = w2 + 2 = 1600 16 = h = 9w 16 w2 + h2 = 1600 er enormups site to ono lessl is grow anousups lo malave . 16) 202+ camisups worry will to you to esidenay adi lo one right 256 =1600 masih unian bavloa ~ 34.86 bn 191 19(34.86)and smile out in enonups novig and to the 16 1619.61 0() Therefore, a 40-inch TV with aspect ratio 16 : 9 is about 35.86 inches wide and 19.61 inches high. Example 1.6.7 Myra purchased a small 25" TV for her kitchen. The size of a TV is measured on the diagonal of the screen. The screen also has an area of 300 square inches.What are the length and width of the TV screen?Written Work/s: Activity 1:Use the three methods (Elimination, Substitution, Graphical Method) to determine the solution(s) of the system of equations below. 1 . x'- x-y =11 X-y = 2 2 . x2 + yz = 1 x 2 - yz = 2 3. Performance Task/s: Activity 2: Read and analyze the problem carefully and do as required. 1. A small television has a picture with a diagonal measure of 10 inches and a viewing area of 48 square inches. Find the length and width of the screen
