Republic of the Philippines Department of Education Computer Science High School of Bicolandia San Jose, Pili, Camarines Sur
Detailed Lesson Plan in Algebra for Grade 8 March 18, 2016 !
"B#EC$%ES At the end of the lesson, students s tudents must be able to: 1. efine efine s!stem s!stem of linear linear e"uati e"uations ons.. 2. #ra$h #ra$h a %i&e %i&enn s!stem s!stem of line linear ar e"uat e"uation ions. s. '. Classif! Classif! (hether (hether a %i&en s!stem s!stem of linear linear e"uation e"uationss is consistent, consistent, inconsis inconsistent, tent, or de$end de$endent. ent. ). *elate *elate the conce$t conce$t of of s!stem s!stem of linear linear e"uations e"uations into into real real life life settin%. settin%.
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S&B#EC$ 'A$$ER A. +o$ic: +!$es +!$es of of s!stem s!stem of linear linear e"uations e"uations . *efere *eference nces: s: -nterm -ntermedia ediate te Al%ebra Al%ebra:: A boo boo for second second !ear/ !ear/ Soleda Soledadd Josei Joseilao lao and Juliet Julietaa #. ernabe/ Pa%e 11 and #oo%le C. Materials: Materials: Cutouts, Cutouts, acti&i acti&it! t! sheets sheets and slide $resen $resentatio tationn . Sills Sills to e&elo$: e&elo$: Anal!sis, Anal!sis, critical critical thinin thinin%% . Conce$t: Conce$t: +(o +(o or more linear linear e"uations e"uations considered considered to%ether to%ether form a s!stem s!stem of linear linear e"uations. e"uations. 3. 4alues -nte%ratio -nte%ration: n: Camarad Camaraderie, erie, team (or, selfesteem selfesteem #. Strate%!: Strate%!: )A5s )A5s Acti&it! Acti&it!,, Anal!sis, Anal!sis, Abstract Abstraction ion and A$$lication A$$lication77
! PR"CED&RE $eacher(s Hint $eacher (s Acti)it* Preliminar* Acti)it* 1. #reetin%s #ood mornin% class 2.
'.
Pra!er
Securin% Cleanliness
#ood mornin% Ma5am Sal&e
Please stand for the o$enin% $ra!er. ;;;;;;;; Students (ill do7 $lease lead the o$enin% $ra!er.
of
efore !ou tae !our seats, indl! $ic u$ the $ieces of $a$er and $lastic or an!thin% that is not a$$ealin% a$$ealin% to the human e!e. +hen, arran%e !our chairs accordin%l! and seat $ro$erl!.
).
Student(s Acti)it*
C h e c i n% Attendance
of
.
C h e c i n% Assi%nment
of
6.
*ecall of th the Pre Pre re"uisite 9esson
o (e ha&e an assi%nment from the $re&ious lesson (e had= Since (e don5t ha&e an assi%nment, let us >ust reca$ linear e"uation. Can an!one tell me (hat a linear e"uation is=
Students (ill do7
eadle re$orts the names of the absentees7
Bone ma5am
9inear e"uation is an al%ebraic e"uation in (hich each term has an e$onent of one.
+he %ra$h of linear e"uation is a strai%ht line. 4er! %ood ?o( about its %ra$h= Bone ma5am Precisel! o ! ou "ues "uestition ons@ s@cl clar arifific icat atio ions ns e"uation=
still h a& e a n! re%a re%ard rdin in%% line linear ar
-f !ou no lon%er ha&e a "uestion, let us no( $roceed to our ne( lesson. efore (e $roceed to the main landmar of our tour called ne( learnin%5, let us first tae a loo
at the follo(in% $ictures. 1st $icture .
Moti&ation
Ees ma5am -t is the maret in Ba%a cit!. Are !ou familiar (ith the $lace in this $icture=
Ma5am, -5&e been into that $lace.
Ma5am +he roads intersect.
+hat (as nice Bo(, let us consider the roads in this this $icture $icture..
Ees ma5am -t is also in the maret of Ba%a cit!. cit!. ?o( about this $icture= Are !ou also familiar (ith the $lace in this $icture= $icture=
+he t(o roads are $arallel to each other.
#ood Since !ou are familiar (ith this $lace, let us tae a loo at the roads in this $icture.
Ees ma5am ?o( about the last $icture= Are !ou also familiar (ith the $lace in this $icture= $icture= 4er! %ood +his is the s!(a! of S9D Alaban%. 9et us also consider the roads in this $icture. As !ou can see there is a s!(a! and a main (a! that is basicall! under the s!(a!. Can !ou see it=
Ees ma5am
Bo ma5am
ased from the $icture, the s!(a! and the main
-f !ou can5t see the main (a!, (hat do !ou thin no( is the relationshi$ of the s!(a! and the main (a! that is under the s!(a!= 4er! %ood Class, in al%ebr al%ebraa (e ha&e ha&e these these (ords that can used to name the roads that inters intersect ect,, the roads that that are $arallel $arallel and the roads that coincide.
B! Less Lesson on Pro Proper per
1.
Presentation the 9esson
of
2.
Presentation of the 9esson Fb>ecti&es
+he (ords that (e (ill use to name the roads that intersect, $arallel and coincide are basicall! our concern for toda!. ecause toda!, (e (ill be dealin% (ith the three t!$es of s!stem of linear e"uations.
At the end of the lesson, students must: 1. efine efine s!stem s!stem of linear linear e"uati e"uations ons.. #ra$h a %i&e %i&enn s!stem s!stem of line linear ar e"uat e"uation ions. s. 3or us to be %uided (ith our lesson, (e ha&e 2. #ra$h '. Clas Cl assi sif! f! (heth (he ther er a %i&e %i &en n s!st s! stem em of line li near ar here the follo(in% ob>ecti&es that (e need to e"uati e"uations ons is consis consisten tentt and inde$e inde$ende ndent, nt, attain attain durin% durin% and after after the discuss discussion ion of the inconsistent, or de$endent. to$i to$ic. c. Can Can an!o an!one ne &olu &olunt ntee eerr to read read thes thesee ). A$$l! the conce$t of s!stem of linear ob>ecti&es= e"uations into real life settin%.
+han !ou 3or us to understand the thin%s that (e (ill disc discus usss toda toda!!, (e need need first first to unlo unloc c the the follo(in% terminolo%ies: Consistent
'.
G n l oc i n % difficulties
of
A stud studen entt rais raises es a hand hand,, (as (as call called ed and and reco%niHed, then reads the ob>ecti&es7
-t is a t!$es of linear s!stem that has eactl!
one solution and (hose %ra$h consists of intersectin% lines. nconsistent -t is a t!$e of linear s!stem that has no solu solutition on and and (hos (hosee %ra$ %ra$hh cons consis ists ts of $arallel lines. Dependent -t is also a t!$e of linear s!stem that has infi infini nite te numb number er of solu solutition onss and and (hos (hosee %ra$h consists of coincidin% lines. Solution -t is a $oint (here the t(o lines intersect@meet. efo efore re (e disc discus usss an!t an!thi hin% n% abou aboutt our our ne( ne(
lesson, let us ha&e first an acti&it!. acti&it!. ).
Acti&it!
a.
PreAc Acti&it!
Mechanics: 1. +he +he clas classs (ill (ill be di&i di&idded into into thre threee %rou$s. 2. ach ach %rou %rou$$ (ill (ill be %i&e %i&enn t(o e"u e"uat atio ions ns to %ra$h. %ra$h. #rou$ 1 (ill (ill %ra$h %ra$h it usin% and ! interce$ts, %rou$ 2 (ill %ra$h it usin% slo$einterce$ slo$einterce$tt form and %rou$ ' (ill %ra$h it usin% table of &alues. '. ach %rou$ (ill %ra$h the t(o e"uations in one Cartesian $lane for minutes onl!. ) . ac h %rou$ (ill c h o os e a re$resentati&e to $resent their (or. 3or !ou to ha&e a hi%h score in the acti&it!, $lease be %uided b! the follo(in% rubrics.
Students count from 1'7 Students (ill do7
Students (ill (or the acti&it! out7
Bo(, $lease start countin%. Proceed no( to !our res$ecti&e %rou$s and start the acti&it! immediatel!. immediatel!. +eacher +eacher facilitates facilitates the acti&it! acti&it!.. She (ill roam around to monitor student5s (ors7 b.
Acti&it! Pro$er
"uations %i&en to each %rou$: #rou$ 1
Students (ill $aste their (ors on the board. &er! re$resentati&e $resents their (ors7 Solution and #ra$h of #rou$ 17
x + y =6
•
− x + y = 4
#rou$ 2 x + y =6
•
x + y = 4 #rou$ ' •
c.
Post Ac Acti&it &it!
2 x − y =1 4 x −2 y =2
+ime5s u$ Please $ost !our (ors on the board
Solution and #ra$h of #rou$ 27
and $resent it. #rou$ 1, $resent !our (or on the board.
Solution and #ra$h of #rou$ '7 #ood >ob #rou$ 1 9et us no( $roceed to the $resentation of %rou$ 2.
4er! er! %ood %ood %rou %rou$$ 2. Bo( Bo(, let let us hear hear the the $resentation of the last %rou$.
ased from our out$uts, out$uts, - thin thin s!stem of linear e"uati e"uations ons is the com$osit com$osition ion of t(o or more more e"uations (orin% to%ether.
Ma5am, the (or of %rou$ 1.
cellent out$uts class 9et us cla$ our hands for oursel&es. .
Anal!sis
ase asedd from from !our !our out$ out$ut uts, s, (ho (ho can can defi define ne a s!stem of linear e"uations=
+he (or or of %ro %rou$ 1 is an eam eam$$le of a consistent s!stem because the %ra$h consists of intersectin% lines. Since the %ra$h is intersectin% lines, - thin it has onl! one solution.
4er! %ood S!stem of linear e"uations has three t!$es: consistent, inconsistent, inconsistent, and de$endent. de$endent.
- thin the (or of %rou$ 2 is an eam$le of an inconsistent s!stem because the %ra$h consists of $arallel lines. -t has no solution.
rilliant ?o( about its solution= oes consistent s!stem has one solution, no solution or infinitel! man! solutions=
-t has no solution since the lines did not intersect at a certain $oint.
cellent -ts solution is the $oint (here the t(o lines intersect.
- thin it is an eam$le e am$le of a de$endent s!stem.
?o( about inconsistent s!stem=
Since the %ra$h consists of coincidin% lines, it made made me thin thin that that it is an eam eam$l $lee of a de$endent s!stem.
4er! %ood -f the %ra$h of an inconsistent s!stem is $arallel, (hat do !ou thin (ith its solution=
- thin it has infinite number of solutions because the t(o lines coincide.
4er! %ood ?o( about the (or of %rou$ '=
Bone Ma5am
Precisel!
rilliant ?o( about its solution=
4er! %ood %ood Al(a!s Al(a!s rememb remember er class class that that the $oint@s (here the t(o lines met is@are actuall! the solution@s of the s!stem. o !ou !ou stil stilll ha&e ha&e an! an! "ues "uestition onss re%a re%ard rdin% in% consistent, inconsistent and de$endent s!stem= 6.
Abstraction
9et us continue our discussion. asicall!, before (e can sa! that a $articular s!stem is consistent, inconsistent or de$endent, there are conditions that must be satisf! first. A s!stem can onl! be said as consistent s!stem if it satisfies these three conditions: 1. +he +he %ra$h %ra$h should should consi consist st of inter intersec sectitin% n% lines. 2. +he s!stem s!stem shou should ld ha&e ha&e onl! onl! one one solut solution ion '. +he s!stem should satisf! the first $ro$ ert!
a1 of linear s!stem,
Ees ma5am
Ees ma5am -t has onl! one solution.
b1 a1
a2 I b2
9et us tae a loo at the linear s!stem %i&en to %rou$ 1. 9et us see (hether the s!stem satisfies the three conditions.
x + y =6 x + y = 4
−
oes oes the the %ra$ %ra$hh of the the s!st s!stem em cons consis istt of intersectin% lines= Since Since the %ra$h %ra$h inters intersect ects, s, then then the s!stem s!stem alread! satisfies the first condition. ?o( about the second condition= oes the s!stem ha&e onl! one solution= actl! ?o( about the last condition=
b1
a2 I b2
1
−1
−1
≠
1 1
≠1
Ees ma5am Since the three conditions ha&e been satisfied, then (e can no( sa! that the s!stem %i&en to %rou$ 1 is reall! a consistent one. Just lie consistent s!stem, inconsistent s!stem has also these three conditions:
+he s!stem also satisfies the second condition since it has no solution because the %ra$h is $arallel. Ees ma5am
1. +he %ra$h %ra$h shoul shouldd consis consists ts of $arall $arallel el line liness 2. +he s!stem s!stem should should ha&e ha&e no no solu solutio tionn '. +he s!stem should satisf! the second
a1 $ro$ert! of linear s!stem,
a1
b1
c1
a2 b2 I c2
b1
1
a2 b2
1
c1
1
I c2
1
1
1
6
I
4 3
I
2
9et us tae a loo at the s!stem %i&en to %rou$ 2 and and let let us see see (het (hethe herr it sati satisf sfie iess the the ' conditions of an inconsistent s!stem.
x + y = 4 oes the s!stem consist of $arallel lines= +he s!stem alread! satisfies the first condition. ?o( about the second condition=
4er! %ood ?o( about the last condition= o !ou thin it (ill be satisf!= Ees ma5am 9et us see.
+he first and the second condition. Since the three conditions ha&e been satisfied, (e can no( sa! that the s!stem %i&en to %rou$ 2 is reall! an inconsistent s!stem.
a1
c1
a2 b2 c2
9et 9et us cont contin inue ue.. Just Just lie lie cons consis iste tent nt and and inconsistent s!stem, de$endent s!stem also has these three conditions to be satisfied in order for a %i&en s!stem to be considered as de$endent s!stem.
2 4
−1
1 2
?ere are the conditions: 1. +he +he %ra$h %ra$h should should consi consists sts of coin coinci cidi din% n% lines 2. +he s!stem s!stem should should ha&e ha&e infini infinite te numb number er of of
b1
Bone ma5am
−2
1
1
2
2 1
2
solutions '. +he s!stem should satisf! the third
a1 $ro$ert! of linear s!stem,
b1
a2 b2
c1
c2 .
A$$lication 9et us tae a loo at the s!stem %i&en to %rou$ '. Are there conditions that are satisfied alread!=
4 x −2 y =2
Ees ma5am 1. Cons Consis iste tent nt s!st s!stem em 2. -nco -ncons nsis iste tent nt s!st s!stem em
- learned the three t!$es of s!stems of linear e"uati e"uations ons.. +he consis consisten tent, t, incons inconsist istent ent and de$endent.
8. #en #eneral eraliiHat Hation ion
4er! %ood As !ou can see all of the conditions (ere satisf! b! the s!stem %i&en to %rou$ '. -t means eans that hat it is real reall! l! an eam eam$l $lee of a de$endent s!stem. o !ou still ha&e an! "uestions re%ardin% the conditions that must be satisf! in order for a s!st s!stem em to be cons consid ider ered ed as con consist sisten ent, t, inconsistent or de$endent= +o test (hether !ou reall! understand all the cond condititio ions ns of cons consis iste tent nt,, inco incons nsis iste tent nt or de$endent s!stem, let us ha&e a board (or. -f call !our name, %o to board and %ra$h the %i&en s!stem as (ell as a$$l! the $ro$erties to identif! (hether it5s consistent, inconsistent or de$endent. Are !ou no( read!= 1.
x + y =3
x −2 y =−6
2.
2 x − y =1
2 x − y =−5
Bo(, (hat did !ou learn from our discussion this mornin%=
-n real life, there (ill be $eo$le that (e (ill meet at eactl! one $oint of our li&es then the! %o and (e (ill ne&er see them an!more. -t is >ust lie consistent s!stem.
A(esome 9et us com$lete the table of the three t!$es of s!stem of linear e"uations re%ardin% its +here (ill also be $eo$le that (e (ill ne&er meet number of solution, $ro$ert! and setch of its in our li&es no matter (here (e %o and no matter (hat (e do. -t is >ust lie inconsistent s!stem. %ra$h. ?o(e&er, there (ill be $eo$le that (e (ill meet in our li&es (ho (ill sta! (ith us fore&er no matter (ho and (hat (e are.
4er! %ood ?o( about the table of an inconsistent s!stem=
rilliant Bo(, for the last one, (ho can com$lete the table of a de$endent s!stem=
cellent class Bo(, (ho can relate the conce$t of the the dif differe ferent nt t!$e t!$ess of s!st s!stem em of line linear ar e"uations into real life settin%=
+hat (as (onderful. -ndeed, s!stem of linear e"uations can be used in order to relate the $eo$le that come and $ass into our li&es.
ans(er the follo(in% for 10 minutes.
%!
E%AL&A$"+ A. #ra$h #ra$h the %i&en %i&en s!stems s!stems of linear linear e"uation e"uationss usin% an! method method and a$$l! a$$l! the $ro$erti $ro$erties es to identif! (hether it is consistent, inconsistent, or de$endent. 1.
x + 4 y =2 2 x −8= 7 y
%!
ASSG+'E+$ A. Gse both %ra$hin %ra$hin%% method method and $ro$er $ro$ertie tiess of s!stem s!stem of linear linear e"uati e"uations ons to identif! identif! (hether (hether the %i&en s!stem of linear e"uations is consistent, inconsistent or de$endent. 1.
2 x + 4 y =5 4 x − 8=7 y
2.
4 x = 7−6 y 3
= 2− 2 x
. Stud! ho( to sol&e sol&e s!stem of linear linear e"uations e"uations b! the use use of %ra$hin% %ra$hin% method, substitu substitution tion method method and elimination method.
Pre$ared b!: Sal)acion '! Escarlan Math Student +eacher
Boted b!: 'elcha P! #uminto Coo$eratin% Coo$era tin% +eacher
