Chapter  9:  Solving  Systems  of  Linear  Equations  Algebraically  9.1  Solving  by  Substitution    Solving  systems  by  graphing  might  not  always  be  the  best  option,  perhaps  due  to  the  time  and  accuracy  required.    Solving  systems  algebraically  can  greatly  increase  our  efficiency  in  finding  the  solution.    One  way  to  solve  algebraically  is  to  solve  by  substitution.      Via  this  method,  you  must:  

1. Label  the  equations  (1)  and  (2)  2. Select  one  of  the  equations  and  pick  a  variable  to  isolate  3. Isolate  the  chosen  variable  in  that  equation  4. Substitute  the  resulting  expression  in  for  that  variable  in  the  other  equation  5. Solve  that  equation  6. Substitute  the  value  found  into  either  equation  to  find  the  value  of  the  other  

variable  7. Check  your  solution  

 Example:  Solve  by  substitution:  2𝑥 − 𝑦 − 1 = 0  4 = 𝑥 + 𝑦                      Example:  Solve  by  substitution:  

𝑦 + 3 =12 𝑥 + 1  

𝑥 − 𝑦 − 2 = 0      

Whenever  this  method  results  in  an  exact  answer  being  found,  the  system  has  exactly  one  solution.    What  if  there  are  infinitely  many  or  no  solutions?    Example:  Solve  by  substitution:  2𝑥 − 3𝑦 + 6 = 0  

𝑦 − 2 =23 (𝑥 + 3)  

             Example:  Solve  by  substitution:  

𝑦 − 2 =23 𝑥 + 3  

𝑦 =23 𝑥 + 4  

               In  word  problems,  you  will  have  to  determine  the  equations  and  variables.    Example:  At  a  particular  grocery  store,  apples  cost  3  times  as  much  as  bananas.    The  prices  are  still  decent,  as  one  can  still  get  a  dozen  bananas  and  10  apples  for  $10.50.    What  is  the  unit  price  for  each?                    Homework:  Pages  474-­‐479,  #1,  2,  4a,  6,  11,  20  

Chapter  9:  Solving  Systems  of  Linear  Equations  Algebraically  9.2  Solving  by  Elimination    Solving  systems  by  substitution  works  very  well  quite  often,  but  it  is  not  the  only  way  to  solve  systems  algebraically.    There  are  a  few  factors  that  can  make  solving  by  substitution  unappealing,  and  we  should  be  familiar  with  multiple  methods.    Solving  by  elimination  involves  applying  linear  transformations  to  the  equations  so  that  we  can  combine  them  to  eliminate  a  variable.    The  linear  transformation  we  will  use  here  will  be  the  multiplication  of  each  side  of  an  equation  by  a  constant.    To  solve  by  elimination,  you  must:  

1. Label  the  equations  (1)  and  (2)  2. Select  a  variable  to  eliminate  3. Rearrange  the  equation  to  have  variables  organized  on  one  side  and  the  constant  

term  on  the  other  4. Multiply  each  equation  by  a  constant  so  that  the  coefficients  for  that  variable  are  

additive  opposites  (if  we  add  them  together,  the  result  is  0)  5. Stack  the  two  equations  and  add  them  together  6. With  one  variable  eliminated,  solve  for  the  other  7. Solve  for  the  missing  variable  as  usual  8. Check  your  solution  

 Example:  Solve  by  elimination:  2𝑥 − 𝑦 − 1 = 0  4 = 𝑥 + 𝑦      

Example:  Solve  by  elimination:  2𝑥 − 3𝑦 + 6 = 0  

𝑦 − 7 = −32 (𝑥 − 1)  

                         Example:  Justin  and  Ryan  go  shopping  for  protein  bars  and  jugs  of  milk.    Justin  buys  20  protein  bars  and  2  jugs  of  milk  for  $63.30.    Ryan  buys  27  protein  bars  and  3  jugs  of  milk  for  $86.70.    Mr.  Athayde  hopes  these  items  last  them  quite  a  while.    What  is  the  cost  for  each  protein  bar  and  jug  of  milk?                                            Homework:  Pages  488-­‐491,  #1,  4,  5a,  6,  7,  13,  15,  18  

Chapter  9:  Solving  Systems  of  Linear  Equations  Algebraically  Supplemental:  Solving  Systems  in  Three  Variables    We  have  explored  relationships  between  two  parameters,  but  there  are  cases  where  three  parameters  are  connected.    We  will  look  at  solving  linear  systems  of  equations  that  involve  three  variables  using  the  process  of  elimination.    Example:  Solve  the  system.  2𝑥 + 𝑦 + 𝑧 = 2    𝑥 − 𝑦 − 3𝑧 = 2    3𝑥 + 2𝑦 + 𝑧 = 1      Firstly,  we  should  number  the  equations  for  easy  reference.    Next,  let  us  choose  a  variable  to  eliminate.    We  will  combine  two  of  the  equations  in  such  a  way  that  the  variable  chosen  is  eliminated,  which  may  involve  multiplying  the  equations  by  a  constant  value.                We  now  have  an  equation  that  involves  only  two  variables.    We  will  then  create  another  by  choosing  another  pair  of  equations  and  eliminating  the  same  variable  as  before.                Now  we  have  a  system  of  two  equations  with  two  variables,  and  we  can  solve  this  using  either  substitution  or  elimination.            Now  that  we  have  a  value  for  one  variable,  we  can  find  the  values  of  the  others  using  equations  we  developed  or  equations  originally  given,  and  state  the  final  answer.          

How  can  we  ensure  the  correct  solution  has  been  attained?            What  might  we  look  for  when  choosing  which  variable  to  eliminate?                Note  that  we  have  three  options  for  variables  to  eliminate,  and  once  that  has  been  chosen,  there  are  three  possibilities  for  pairs  of  equations  to  combine  to  eliminate  that  variable.    There  are  options  galore!    Some  options  might  be  easier  to  work  with  than  others,  but  all  will  lead  to  the  same  answer.    Example:  Solve  the  system.  2𝑥 − 𝑦 + 𝑧 = 1    −3𝑥 + 2𝑦 − 𝑧 = 2    4𝑥 + 3𝑦 − 3𝑧 = −3        

Example:  Solve  the  system.  𝑥 + 2𝑦 + 3𝑧 = 8    2𝑥 − 𝑦 + 2𝑧 = 0    5𝑥 + 3𝑦 − 𝑧 = 2                                  Example:   There   are   3   lower-­‐case-­‐h   hardcore   gamers   getting   some   new   hardware   and  games.    One  buys  a  console,  3  games,  and  3  controllers  for  a  total  of  $500.    The  second  buys  a   console,   7   games,   and   1   controller   for   a   total   of   $660.     The   last   one   is   slightly   more  hardcore,   but   is   crazy   rich   and   can   afford   it.     He   buys   3   consoles   (so   he   can   have   LAN  parties),  18  games,  and  9  controllers  for  a  total  of  $2040.    Find:  

• The  cost  of  a  console  • The  cost  of  a  game  • The  cost  of  a  controller  

   

Supplemental  Section  Homework:    Practise  

1. Which  of  the  given  ordered  triples,  if  either,  is  a  solution  of  the  given  equation?  a. 𝑥 + 𝑦 − 𝑧 = 2; 1,−2,−3 , (0,−2,−4)  b. 𝑥 − 𝑦 + 2𝑧 = 3; 1,−1,−1 , (2,−3,−1)  c. 𝑥 + 4𝑦 + 2𝑧 = 0; 0, 0, 4 , (0,−2, 8)  d. 5𝑥 + 𝑦 − 𝑧 = 0; −1,−1, 6 , (2,−6, 4)  

 2. Evaluate  the  following  expression:  

2𝑥 + 3𝑦 + 3𝑧  when  𝑥 = 2,𝑦 = −2, 𝑧 = 5    

3. Which  values  should  be  given  to  𝑎, 𝑏,  and  𝑐  so  that  the  linear  system  has   −1, 2,−3  as  its  only  solution?  

𝑥 + 2𝑦 − 3𝑧 = 𝑎    −𝑥 − 𝑦 + 𝑧 = 𝑏    2𝑥 + 3𝑦 − 2𝑧 = 𝑐    

 4. Solve  the  following  systems:  

a. 3𝑥 + 2𝑦 − 2𝑧 = 11  𝑥 + 𝑦 + 2𝑧 = −1    2𝑥 − 𝑦 + 2𝑧 = −4      

b. 3𝑥 − 4𝑦 + 5𝑧 = 2  4𝑥 + 5𝑦 − 3𝑧 = −5    5𝑥 − 3𝑦 + 2𝑧 = −11      

c. 𝑥 + 𝑦 + 3𝑧 = 12  2𝑥 + 𝑦 + 3𝑧 = 13    𝑥 − 𝑦 + 4𝑧 = 11            

Apply  5. The  equation  of  a  circle  may  be  written  as  𝑥! + 𝑦! + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0.    The  

coordinates  of  any  point  on  the  circle  must  satisfy  its  equation.  Find  the  values  of  𝐷,𝐸,  and  𝐹  so  that  the  circle  passes  through  (1,  1),  (-­‐2,  3),  and  (3,  4),  and  write  the  equation  for  the  circle.  

 6. Find  𝐴,𝐵,  and  𝐶  so  that  the  solution  set  of  the  equation  𝐴𝑥 + 𝐵𝑦 + 𝐶𝑧 = 1  will  

contain  the  given  ordered  triples:  (4,  1,  2),  (3,  2,  1),  (-­‐6,  -­‐1,  -­‐2)      

Answers:    

1. a.    Both  b.     2, 3,−1  c.    Neither  d.     2,−6, 4  

2. 13  3. 𝑎 = 12, 𝑏 = −4, 𝑐 = 10  

 

4. a.     𝑥,𝑦, 𝑧 = 1, 2,−2  b.     𝑥,𝑦, 𝑧 = −2, 3, 4  c.     𝑥,𝑦, 𝑧 = 1, 2, 3  

5. 𝐷 = −1,𝐸 = −7,𝐹 = 6;    𝑥! +𝑦! − 𝑥 − 7𝑦 + 6 = 0  

6. 𝐴 = −1,𝐵 = 1,𝐶 = 2

Chapter  9:  Solving  Systems  of  Linear  Equations  Algebraically  Review    Textbook  review:  Pages  502-­‐503,  #1-­‐13    Additional  questions:  

1. Solve  the  system:  𝑥 + 2𝑦 − 𝑧 = 1  2𝑥 − 5𝑦 + 𝑧 = 17  2𝑥 + 6𝑦 − 3𝑧 = −2  

2. Solve  the  system:  2𝑥 + 4𝑦 − 3𝑧 = 3  𝑥 − 4𝑦 − 𝑧 = 2  

−2𝑥 + 2𝑦 + 5𝑧 = −1  3. Solve  the  system:  

2𝑥 + 4𝑦 − 𝑧 = 1  2𝑥 + 𝑦 − 2𝑧 = 1  3𝑥 + 9𝑦 − 3𝑧 = 4  

4. In   a   particular   badminton  match,   the   time   of   each   game  was   divided   between  rallies,  time  between  rallies,  and  the  mid-­‐game  interval.    In  the  first  game,  there  were  35  rallies,  34  periods  between  rallies,  and  1  mid-­‐game  interval  for  a  total  of  679   seconds.     In   the   second   game,   there  were   25   rallies,   24   periods   between  rallies,   and  1  mid-­‐game   interval   for   a   total   of   499   seconds.     In   the   final   game,  there  were  56  rallies,  55  periods  between  rallies,  and  2   intervals  granted   for  a  total   of   1117   seconds.     Find   the   average   length  of   each   rally,   the   average   time  between  each  rally,  and  the  length  of  the  interval.