III-3 Systems of Equtions Multiple Equtions with Multiple Unknowns: The generl rule tht you need to be wre of is tht to solve for two unknowns, you need two independent equtions contining those two unknowns (nd no other unknowns. More generlly, you need n independent equtions to solve for n unknowns (where n is ny number ny positive integer, to be precise. The techniques for solving for the unknowns re illustrted below. Independent Equtions: If you cn obtin one (or more of your equtions by lgebriclly combining the other equtions, the set is not independent. This is not lwys esy to see, but it is importnt to be wre of. It will certinly come up in your physics courses. Exmple 1: Is this set of equtions independent or not? Tht is, cn you get one of these equtions s combintion of the other two equtions? (1 A + B = C ( A B = D (3 A = C + D This set is NOT independent. If you dd eqution (1 nd eqution (, the result is eqution (3. Thus, eqution (3 provides no informtion beyond wht is given by equtions (1 nd ( nd you cnnot solve for three unknown vribles using this set of equtions. In fct, you should show tht combining ny two equtions yields the third. Any two of these equtions do mke n independent set, so you could solve for two unknowns. Exercise 1: For ech set of given equtions, how mny independent equtions re there? How mny unknowns could you solve for? ( (1 A + B = C ( A B = D (3 C + 3 D = 7 (b (1 V 1 I 1 R 1 I R = ( V 1 I 1 R 1 I 3 R 3 = (3 I 3 R 3 I R = Solving Multiple Equtions: The generl pln is to lgebriclly combine the equtions so tht you end up with one eqution with only one unknown. You then solve for tht unknown nd use it to work bck through your other equtions to find the other unknowns. The two generl methods for doing this re substitution nd elimintion. Generl Procedure for Substitution Method 1. Solve one eqution for one unknown in terms of the other(s.. Then substitute the result into nother eqution nd solve it for the next unknown. 3. Continue until you hve n expression for one unknown in terms of known quntities. 4. Then work bckwrds to find the other unknowns. Generl Procedure for Elimintion Method 1. Add one eqution or multiple of one eqution to second eqution to eliminte one of the unknowns.. Repet this process until you hve one eqution in one unknown nd solve for tht unknown. 3. It is lso sometimes useful to divide one eqution by nother to eliminte one of the unknowns. CSM Physics Deprtment 1998-1
III-4 Let s look t few exmples tht illustrte the methods tht you will be using. We will first solve system of equtions by substitution to illustrte tht method. For simplicity, we will first work with two independent equtions nd two unknowns. Exmple : Solve the following system of equtions for the unknowns B nd C: (1 A + B C = D ( B + C = D Note tht A nd D re considered known quntities. Step 1: Solve for one of the unknowns in terms of the other from either eqution. With prctice, you lern to spot wht might be the esiest pproch. In this cse, we cn solve for either B or C from either eqution. For this exmple, let s use eqution ( to get n expression for B. Subtrct C from ech side of eqution ( nd obtin: B = D C Step : Replce B in eqution (1 with D C. The result will be n eqution with only C s n unknown. A + B C = D becomes A + (D C C = D which is A 3C = D Now isolte the unknown, C: And solve! A D = 3C C = (A D/3 Since A nd D re known, now C is now known. Step 3: Let s find B by substituting this expression for C bck into either of the originl equtions or the eqution we found t the end of Step 1. At the end of Step 1, we found B = D C. Substituting the expression in for C we obtin B = D [(A D/3] = D 3 A + 3 D, so B = 5 3 D 3 A (Notice tht the - when distributed gve us +/3 D. We cn now report the solution s 5 A D B = D A nd C =. 3 3 3 The system of equtions hs now been solved. We were told tht we lredy knew A nd D, so we found expressions for B nd C in terms of them. Any correct lgebric mnipultions of the equtions would hve gotten us to the exct sme results. But don t tke my word for it, check it out yourself! Exercise : For the sme two equtions, (1 A + B C = D nd ( B + C = D, solve for B nd C using the following steps. Step 1: From eqution (1, find n expression for C. Step : Substitute the expression for C found in step one in eqution (. Step 3: Solve for B using the new form of eqution (. Step 4: Substitute the expression for B found in step 3 into either originl eqution nd solve for C. Step 5: Compre your results to those obtined in the exmple. If they re not the sme, one of us mde mistke. Exercise 3: For the sme two equtions, solve for B nd C by using the following steps. CSM Physics Deprtment 1998-1
III-5 Step 1: From eqution (, find n expression for C. Step : Substitute the expression found in step one for C in eqution (1. Step 3: Solve for B using the new form of eqution (1. Step 4: Substitute the expression for B found in step 3 into either originl eqution nd solve for C. Step 5: Compre your results to those obtined previously. Exmple nd Exercises 3 nd 4 should illustrte for you tht no mtter which pproch you choose, ll the substitution methods yield the sme solution. Now tht you ve seen nd prcticed substitution, let s look t the elimintion method. We will continue to work with the sme system of equtions nd the sme unknowns. Exmple 3: Solve the following system of equtions for the unknowns B nd C: (1 A + B C = D ( B + C = D Step 1: To get rid of B, subtrct eqution ( from eqution (1. This is the sme s multiplying eqution ( by negtive 1nd then dding the equtions. (1 1 ( Sum A + B C = D B C = D A 3C = D Now we hve n eqution which does not contin the unknown B. We hve eliminted B. Step : Solve for C. It is esy to see tht we obtin the sme result s we did before. C = (A D/3 Step 3: As before, substitute this expression into either eqution nd find B. The result, of course, will be identicl to tht obtined in ll of the previous solutions. Oky, your turn. Exercise 4: Solve the sme system of equtions for B nd C. Step 1: Add multiple of eqution (1 to eqution ( to eliminte C. Step : Solve the resulting eqution for B. Step 3: Substitute your expression for B bck into eqution (1 or ( nd solve for C. The equtions we encounter in physics re not lwys liner. The next exmple illustrtes both methods pplied when one eqution contins qudrtic term with n unknown. Exmple 4: Solve x = x + v t + ½ t nd v = v + t for v nd t. Substitution: We could solve either eqution for the unknown v or the unknown t. In generl, we will wnt to choose to solve the simpler eqution (in this cse not the qudrtic! for one unknown. If we solve the second eqution for t, we will hve to substitute it into two plces in the first eqution nd we would hve to squre it when substituting into the ½ t term. If we solve insted for v, we will only hve to substitute it into one plce in the first eqution (nd not squre it!. Here we go! CSM Physics Deprtment 1998-1
III-6 1. Solve the second eqution for v. v = v t This cn be done in 1 lgebric step! Wht lgebric step ws required?. Substitute the result into the first eqution. x = x + (v t t + ½ t We now hve n eqution with only one unknown! 3. Solve for t: x = x + v t t + ½ t = x + v t ½ t x x v t + ½ t = - t = (- v ± v - 4( 1 ( x - x ( 1 v ± = v - ( x - x 4. Substitute the result for t bck into one of the originl equtions or the eqution from step 1. v v - ( x - x v v - ± = = v - ( v ± v - ( x - x = m v - ( x - x The solution cn now be reported s t v v - ( x - x = nd v + v - ( x - t v + v - ( x - x = nd v v - ( x - x =. Note, we hve to correctly pir up our vribles ( x - v = + is NOT solution. v - x Elimintion: 1. Multiply the second eqution by -t. v t = v t t. Add the result to the first eqution. x = x + v t + ½ t v t = v t t Sum: x v t = x ½ t Notice tht the v t nd the v t cncelled. Agin, we hve n eqution with only one unknown! = or x ( x - x v + v - t = nd 3. Solve for t: x v t = x ½ t x x v t + ½ t = This is the exct sme eqution we pplied the qudrtic formul to solve in the previous method. From here the steps re the sme s the substitution method. After you hve t, substitute bck into one of the originl equtions nd solve for v CSM Physics Deprtment 1998-1
III-7 Now try your hnd t these. Pick whichever method you like. If you re feeling mbitious, try few different pproches for ech. Note tht those vribles not identified s unknown re to be treted s known. Exercise 5: Solve the system of equtions (1λ + μ = 11ω nd ( 4μ + ω = 7λ for λ nd μ. (Which vrible is considered known? Exercise 6: Solve the system of equtions (1 y = x + nd ( y = 8x 1 for x nd y. Once you hve solved for x nd y lgebriclly, crefully sketch the curves for ech eqution on the sme grph. Find the point t which the curves intersect nd compre with your lgebric result. OK, the next exercise is here to illustrte specil cse tht sometimes comes up when solving systems of equtions. Exercise 7: Solve the system of equtions (1 x y = nd ( y = x + for x nd y. First, see wht hppens when you solve lgebriclly by substitution or elimintion. Wht does this men? Cn these two equtions both be true t the sme time? After you hve considered the problem lgebriclly, grph ech eqution nd see if your lgebric findings mke more sense to you. So fr the exercises you ve been sked to do hve turned out firly simple. Rel world problems re not lwys simple. Exercise 8: Solve the system of equtions (1 p t 4r = 7 nd ( t 1 = r + 3 for r nd t. Use substitution or elimintion to solve. Here is wht you should find: 7 4 p r = nd p 4 9 t =. p 8 Now, this is not the only wy the solutions cn be written, if you get something different, check tht you get these expressions if you write r nd t s single frction. CSM Physics Deprtment 1998-1