NCTM Ntionl Conference, 2017 Arithmetic & Algebr Hether Dlls, UCLA Mthemtics & The Curtis Center Roger Howe, Yle Mthemtics & Texs A & M School of Eduction
Relted Common Core Stndrds First instnce of vrible : CCSS.MATH.CONTENT.1.OA.A.1 Use ddition nd subtrction within 20 to solve word problems involving situtions of dding to, tking from, putting together, tking prt, nd compring, with unknowns in ll positions, e.g., by using objects, drwings, nd equtions with symbol for the unknown number to represent the problem.1
Relted Common Core Stndrds First forml instnce of vrible: CCSS.MATH.CONTENT.3.OA.D.8 Solve two-step word problems using the four opertions. Represent these problems using equtions with letter stnding for the unknown quntity. Assess the resonbleness of nswers using mentl computtion nd estimtion strtegies including rounding.
Relted Common Core Stndrds First forml solving of equtions: CCSS.MATH.CONTENT.6.EE.B.5 Understnd solving n eqution or inequlity s process of nswering question: which vlues from specified set, if ny, mke the eqution or inequlity true? Use substitution to determine whether given number in specified set mkes n eqution or inequlity true. CCSS.MATH.CONTENT.6.EE.B.6 Use vribles to represent numbers nd write expressions when solving rel-world or mthemticl problem; understnd tht vrible cn represent n unknown number, or, depending on the purpose t hnd, ny number in specified set. CCSS.MATH.CONTENT.6.EE.B.7 Solve rel-world nd mthemticl problems by writing nd solving equtions of the form x + p = q nd px = q for cses in which p, q nd x re ll nonnegtive rtionl numbers.
Connecting Arithmetic & Algebr
Connecting Arithmetic nd Algebr Are the story problems of elementry school nd those we find in lgebr clss relted nd cn they be used to connect lgebr nd rithmetic?
Let s solve with nd without lgebr 1. J. hd some model trucks. Tody he bought four more. Now he hs seven trucks. How mny trucks did he hve? How would we solve this in elementry school? How would we solve this in lgebr clss? Note: This problem is n Add To, Strt Unknown (Tble 1, Appendix A, Common Core Mth Stndrds), which is the most difficult of the Add To types for students to solve.
Vrious Solutions 7 t 4 3 4 If J. ended up with seven trucks fter buying four new trucks, then before he bought the new trucks, he hd 7-4 = 3 trucks. # of trucks he hd # trucks he hs now totls 7? 4 4+4 = 8 no 2 2+4 = 6 no 3 3+4 = 7 yes! Let t be the # of trucks J. hd. Then, t + 4 = 7 nd (t + 4) - 4 = 7-4 t = 3.
How do we wnt to chrcterize our vrious solution methods? One proposl: Digrmmtic Verbl Guess nd Check Algebric
Let s solve with nd without lgebr 2. J. bought three pcks of blloons. He opened them nd counted 12 blloons. How mny blloons re in pck?
Record your work on scrtch s follows:
Let s solve with nd without lgebr 3. J. hs 4 pckges of blloons nd five single blloons. In ll he hs 21 blloons. How mny blloons re in pckge?
Let s solve with nd without lgebr 4. There re 36 children in clss. There re 4 more boys thn girls. How mny boys re in the clss, nd how mny girls? Note: This problem is something of wtershed in the United Sttes, lmost lwys being pproched lgebriclly.
Vrious Solutions: b 36 4 g g 32 4 16 16 g Tke wy four boys. This leves 36-4=32 students, nd n equl number of boys nd girls. Therefore there re 32/2 = 16 girls. # of girls # of boys # of students totls 36? 4 4+4=8 4+8=12 no 10 10+4=14 10+14=24 no 16 16+4=20 16+20=36 yes! Let b be the # of boys in the clss, nd let g be the # of girls. Then, b + g = 36 nd b = g + 4. Solving we hve: b + g = (g+4) + g = 2g + 4. 2g + 4 = 36, 2g = 36-4 = 32, g = 2g/2 = 32/2 = 16.
Let s solve with nd without lgebr 5. If br of sop blnces 3/4 of br of sop nd 3/4 of pound, how much does the br of sop weigh?
Vrious Solutions: Since there is full br of sop on one side of the blnce, nd 3/4 br of sop on the other, we my remove 3/4 of br of sop from ech side of the blnce. The objects remining on ech side will still blnce. Thus 1/4 br of sop weighs 3/4 pounds. A full br of sop is 4 1/4ths, so it weighs 4 x 3/4 = 3 pounds. Let S be the weight of br of sop, in pounds. The weight of one br of sop is S. The weight of 3/4 br of sop nd 3/4 pounds is 3/4 S + 3/4. Hence we hve S = 3/4 S + 3/4. Solving, we hve S - 3/4 S = 1/4 S = 3/4, S = 4 x 1/4 S = 4 x 3/4 = 3.
Vrious Solutions: w 3/4 lb 3 3/4lb 3/4lb 3/4lb 3/4lb weight of one br, in lbs. weight of 3/4 br nd 3/4 lbs. blnced? 1 3/4 x (1lb)+3/4 lb = 1.5 lbs no 1/2 lb 3/4 x (1/2 lb) + 3/4 lb = 9/8 lbs no Note: Creting the second column on the guess nd check tble is non-trivil for mny students.
For ech problem, plese check the two solution methods you believe students would find most ccessible.
Let s solve with nd without lgebr 6. A mn wnts to shre his coins eqully mong some friends. If he gives ech friend 6 coins, he would hve 4 coins left over. If he gives ech friend 7 coins, he would be 5 coins short. With how mny friends does he shre?
Vrious Solutions: f f f f f f 4 f f f f f f f 5 # of friends totl coins if shring 6 ech totl coins if shring 7 ech sme # of coins? 3 6 x 3 + 4 = 22 7 x 3-5 = 16 no 7 6 x 7 + 4 = 48 7 x 7-5 = 44 no 9 6 x 9 + 4 = 58 7 x 9-5 = 58 yes!
Vrious Solutions: How does working with the digrmmtic model become more chllenging when the student must model groups of b objects where is unknown? Consider the chllenge of the digrmmtic model when solving the following problem: H. nd J. re gining weight for footbll. H. weighs 205 pounds nd is gining 2 pounds per week. J. weights 195 pounds nd is gining 3 pounds per week. When will they weigh the sme?
Vrious Solutions: # of friends totl coins if shring 6 ech totl coins if shring 7 ech sme # of coins? 3 6 x 3 + 4 = 22 7 x 3-5 = 16 no 7 6 x 7 + 4 = 48 7 x 7-5 = 44 no 9 6 x 9 + 4 = 58 7 x 9-5 = 58 yes! f 6 f + 4 7 f - 5 6 f + 4 = 7 f - 5 6f - 6f + 4 = 7f - 6f - 5 4 + 5 = 1f - 5 + 5 9 = 1f
Let s solve with nd without lgebr 7. Tickets for the clss show re $3 for students, nd $10 for dults. The uditorium holds 450. The show ws sold out, nd the clss rised $2750 in ticket sles. How mny students bought tickets?
Vrious Solutions: 450 s 2750 s s s 1400 2750-3 x 450 =1400 1400 200 200 200 200 200 200 200 So, 450-200 = 250 students bought tickets Note: In this problem, the br model becomes more chllenging to both drw nd to conceptulize s students del with modeling groups of b where is unknown.
Vrious Solutions: The clss sold 450 tickets. If ll the ticket hd been for dults, totl sles would hve been $4500. Insted, the sles were $2750, which is $4500 - $2750 = $1750 less. Since ech student ticket brings in $10 - $3 = $7 less thn n dult ticket, there must hve been 1750/7 = 250 student tickets sold. Note: This is clled method of flse position. Note how this non-lgebric method requires more ingenuity on this problem compred to the lgebric method. Let s be the number of student tickets sold, n let be the number of dult tickets sold. Then + s = 450 nd 10 + 3s = 2750. + s = 450 = 450 - s 10(450 - s) + 3s = 2750 4500-10s + 3s = 2750 4500-7s = 2750 4500-4500 - 7s = 2750-4500 -7s = -1750 So, s = -1750/-7 = 250 student tickets
Vrious Solutions: # of student tickets income from student tickets income from dult tickets totl income totls $2750? 20 $3 x 20 = $60 $10 x (450-20) = $10 x 430 = $4300 $60 + $4300 = $4360 no 100 $3 x 100 = $300 $10 x (450-100) = $10 x 350 = $3500 $300 + $3500 = $3800 no s $3 x s = 3s $10 x (450 - s) = 10(450 - s) 3s + 10(450 - s) = 2750 3s + 4500-10s = 2750 4500-7s = 2750 7s = 1750 s = 250 3s + 10(450 - s) = 2750
Check the solution method(s) you believe prticulrly id students in ccomplishing the indicted tsk.
Connecting Arithmetic nd Algebr We believe the story problems of elementry school nd those we find in lgebr clss ARE relted nd CAN be used s one of the connections between lgebr nd rithmetic.
Other Avenues for Connecting Arithmetic nd Algebr Vi our nottion for numbers: 156 = 1x10 2 +5 x10 1 +6 x10 0 Vi Proof: Ask students to justify generl clims bout numbers. Exmple: The sum of two odds is even. 156 = 1 x 2 + 5x 1 + 6x 0 = x 2 + 5x + 6 when x = 10 Russell, Schifter & Bstble: http://homepges.wmich.edu/~grntt/ Developing%20Mthemticl%20Arguments.pdf
Other Avenues for Connecting Arithmetic nd Algebr Vi the four opertions: 12 x 13 36 120 156 10 + 2 x 10 + 3 6 30 20 100 156 x + 2 x x + 3 6 3x 2x x 2 x 2 +5x+ 6
Questions For which (if ny) problems ws the lgebric solution method more dvntgeous? If ny, how? Which non-lgebric methods (if ny) prepre students for the lgebric solution method? If ny, how? Wht (if nything) in this tlk helped you think bout using connections between rithmetic nd lgebr in your clssroom?
Contct nd resources dlls@mth.ucl.edu roger.howe@yle.edu Google: UCLA Curtis Center nd click on Ltest Resources for the slides from this tlk & Roger s pper Arithmetic to Algebr