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EDUCATION PROGRAM      FUNDEMENTALS OF ALGEBRA É   ALGEBRA ISº WASTING TIME  ADDING NUMBERS   SOLVING EQUATIONS INVOLVINÇ INDETERMINATEÓ NONE OF THE ABOVE „d‘d‘Z‘d‘x‘ CONGRATULATIONS ! YOU HAVE COMPLETED THE EDUCATION PROGRAM"FUNDAMENTALS OF ALGEBRA I". HOPEFULLY,YOU NOW HAVE SOME IDEA OGO ON) €’ INCORRECT ®ALGEBRA IS THE STUDY OF "EQUATIONS". AÎEQUATION STATES THAT TWO NUMERICAL "QUANTITIES" ARE THE SAME. FOR EXAMPLEº  2 + 2 = 4 0 + 10 = 10   (PRESS RETURN TO GO ON) €´‘ REVIE×  THE MAIN GOAL OF RETURN TO GO ON) €b“ THESE REAL WORLD PROBLEMS ARE USUALLY REFERRED TO AS "WORD PROBLEMS"®HERE IS AN EXAMPLE:  IF JANE WAS TWICE AS OLD AS SHE IÓNOW, SHE WOULD BE 3 YEARS YOUNGER THAÎMARY. MARY IS NOW 25. HOW OLD IS JANE? (PRESS RETURN TO 5X = 1. HERE'S HOW TO DO ITº 1) EXAMINE THE COEFFICIENT OF X  (IN THIS CASE IT IS 5© 2) MULTIPLY BOTH SIDES BY ONE OVEÒ THE COEFFICIENT OF X  (IN THIS CASE 1/5)  3) SIMPLIFY BOTH SIDES AND WE'RE  DONE   (PRESSF THE ABOVÅ„p”‘‘‘z” SORRY ®PLEASE TRY AGAIN® (PRESS RETURN TO GO ON©€z” NO ®THAT IS NOT CORRECT. PLEASE TRY AGAIN® (PRESS RETURN TO GO ON©€f” USING THE MULTIPLICATIVE RULE, WÅCAN SOLVE SIMPLE ALGEBRAIC EQUATIONS LIKE ON) €¦• GREAT ¡REMEMBER ALSO THAT NEGATIVE NUMBERS REP-RESENT THE "ABSENCE" OF A QUANTITY ANÄARE TO THE "LEFT" OF ZERO® (PRESS RETURN TO GO ON©€j• LET'S TRY ANOTHER EQUATION®  (-10) - 2 = ¿  -12   -10 + 2   12   NONE OCE OF A QUAN­TITY. POSITIVE NUMBERS ARE WRITTEN WITHA "PLUS" (+) SIGN IN FRONT OF THEM, OÒNO SIGN AT ALL, BUT NEGATIVE NUMBERÓ"ALWAYS" HAVE A "MINUS" (-) SIGN IÎFRONT OF THEM. FOR EXAMPLE: +4 OR 4 IÓPOSITIVE, AND -4 IS NEGATIVE® (PRESS RETURN TO GOITIVE RULÅ THE MULTIPLICATIVE RULÅ BOTH (A) AND (B© NONE OF THE ABOVÅ„P‘P‘2‘P‘Z‘ THE MOST BASIC NOTION OF ALGEBRÁIS THAT OF "POSITIVE" AND "NEGATIVE¢NUMBERS. POSITIVE NUMBERS SIGNIFY THÅPRESENCE OF A QUANITIY, WHILE NEGATIVÅNUMBERS SIGNIFY THE ABSEN (PRESS RETURN TO GO ON) €€“ REVIEW   AN EQUATION IS EITHER --?-- OR --?--® BIG, SMALÌ POSITIVE, NEGATIVÅ TRUE, FALSÅ NONE OF THE ABOVÅ„ª‘ª‘–‘ª‘´‘ REVIE×  WHICH OF THE FOLLOWING ARE USED TÏSOLVE EQUATIONS¿ THE ADDAME>®PLEASE TRY THE QUESTION AGAIN® (PRESS RETURN TO GO ON©€ò” THAT'S RIGHT ¬SINCE SUBTRACTING A POSITIVE NUMBER IÓTHE SAME AS ADDING A NEGATIVE NUMBER.  (PRESS RETURN TO GO ON©€z” MARVELOUS ¡THAT WAS A DIFFICULT ONE!  LICATIVE RULES® (PRESS RETURN TO GO ON) €‘ THE TRUTH IÓ¬REAL WORLD PROBLEMS ARE USUALLÙCALLED WORD PROBLEMS IN ALGEBRA.  (PRESS RETURN TO GO ON©€ ‘ (-5) + 8 = ¿ -³ ° +³ NONE OF THE ABOVÅ„Ä•Ä•è”Ä•ò” THAT IS NOT CORRECT¬® BOTH THE ADDITIVE "AND" THE MULTI­PLICATIVE RULES ARE USED TO SOLVE EQUA­TIONS.  (PRESS RETURN TO GO ON©€2‘ THAT IS NOT CORRECT¬® IN ORDER TO SOLVE THE EQUATION  5X + 3 = 7 WE NEED BOTH THE ADDITIVÅAND MULTIP RETURN TO GO ON)  €‚‘ SORRY ® AN UNKNOWN QUANTITY IS CALLED AÎINDETERMINATE® (PRESS RETURN TO GO ON)  €x‘ INCORRECT ® THE MAIN GOAL OF ALGEBRA IS TO  SOLVE EQUATIONS INVOLVING INDETERMI­NATES.   (PRESS RETURN TO GO WHILE THE EQUATION 1 = 0 IÓFALSE.   (PRESS RETURN TO GO ON)  €–‘ NOT REALLY ® SUBTRACTING A POSITIVE NUMBER IÓTHE SAME AS ADDING A NEGATIVE NUMBER® SUBTRACTING A NEGATIVE NUMBER IÓTHE SAME AS ADDING A POSITIVE NUMBER® (PRESSDER TO SOLVE THE EQUATIOÎ5X + 3 = 7 WE NEEDº THE MULTIPLICATIVE RULÅ THE ADDITIVE RULÅ BOTH (A) AND (B© NONE OF THE ABOVÅ„(‘(‘‘(‘2‘ INCORRECT ,   AN EQUATION IS EITHER "TRUE" OÒ"FALSE". FOR EXAMPLE THE EQUATION 2 + 2= 4 IS TRUEHIS PROBLEM?  3X + 4 = 2· 3X - 4 = 2· 3X + 27 = ´ 3X - 4 + 27 = °„ú‘ð‘ú‘ú‘’ REVIE×  AN UNKNOWN QUANTITY IS CALLEDº AN INDETERMINATÅ A POSITIVE NUMBER   AN EQUATION   NONE OF THE ABOVE „x‘n‘n‘n‘‚‘ REVIE×  IN ORRULE AND WE HAVE THAT:  (1/2) * 2X = (1/2) * 4 THIS GIVES X = 2.  (PRESS RETURN TO GO ON©€J’ IF JANE WAS THREE TIMES AS OLD AÓSHE IS NOW, THEN SHE WOULD BE 4 YEARÓOLDER THAN MARY. MARY IS 27. HOW OLÄIS JANE? WHICH EQUATION BELOW GOES WITHTTHEN TWO TIMES TWO TAKE AWAY FOUR IS ZERO OR (2*2) - 4 = 0® (PRESS RETURN TO GO ON)  €J’ SORRY ®THE CORRECT ANSWER WAS (A), 2. LET'S SEEWHY:   FIRST APLY THE ADDITIVE RULE: 2X - 4 + 4 = 0 + 4  SO 2X = 4, NOW USE THE MULTIPLICA­TIVE GHT. PLEASE TRY AGAIN® (PRESS RETURN TO GO ON©€¸’ SOLVE: 3X + 2 = -1 X = ¿  -± ± -³ NONE OF THE ABOVÅ„r’æ‘æ‘æ‘|’ NO ¬THAT IS NOT CORRECT. PLEASE TRÙAGAIN® (PRESS RETURN TO GO ON©€|’ CORRECT !  IF X = 2, CATIVE RULE ALONE. TO SOLVÅTHESE EQUATIONS WE NEED TO USE "BOTH" RULES.   FOR EXAMPLE: SOLVE: 5X + 3 = 7  (PRESS RETURN TO GO ON)  €¸’ SOLVE: 2X + 4 = 8 X = ¿  ² ´ ¸ NONE OF THE ABOVÅ„¤’Ò‘Ò‘Ò‘®’ SORRY ¬BUT THAT IS NOT RIAME>¬THAT'S NOT RIGHT. PLEASE TRY AGAIN® (PRESS RETURN TO GO ON©€à’ SO FAR ¬YOU HAVE SEEN TWO DIFFERENT RULES WHICÈARE USED TO SOLVE EQUATIONS. HOWEVER¬THERE ARE EQUATIONS WHICH CAN NOT BÅSOLVED BY USING EITHER THE ADDITIVE RULEOR MULTIPLI SOLVE: 5X = 3 X = ?   FIRST NOTICE THAT THE COEFFICIENT OF X IS 5, SO MULTIPLY BOTH SIDES OF THEEQUATION BY 1/5. SINCE (1/5) * 5X = ØAND (1/5) * 3 = 3/5, WE HAVE THAÔX = 3/5. PLEASE TRY THE PROBLEM AGAIN® (PRESS RETURN TO GO ON©€“ .  (PRESS RETURN TO GO ON) €:“ INCORRECT ®HERE IS A SIMILAR PROBLEMº  20. BUT (1/20) * 20X = X ANÄ(1/20) * 20 = 1, AND SO WE HAVE THAÔX = 1. PLEASE TRY THE PROBLEM AGAIN® (PRESS RETURN TO GO ON) €X“ THAT IS NOT CORRECT,  ®HERE IS A SIMILAR EXAMPLE:    SOLVE: 3X = 6    X = ?    FIE 1/5 * 5 = 1 AND 1/5 * 1 = 1/5  (PRESS RETURN TO GO ON) €X“ NOT QUITE ®THIS IS A SIMILAR EXAMPLEº  SOLVE: 20X = 20  FIRST NOTICE THE COEFFICIENT OÆX IS 20. SO, MULTIPLY BOTH SIDES OÆTHE EQUATION 1/20, I.E. (1/20) * 20X = (1/20) *® (PRESS RETURN TO GO ON) €v“ LET'S TRY THIS METHOD ON THE EX- AMPLE 5X = 1.  FIRST, EXAMINE THE COEFFICIENT OÆX; IT'S 5. SO MULTIPLY BOTH SIDES OÆTHE EQUATION BY 1/5. THAT ISº (1/5) * 5X = (1/5) * ±SIMPLIFYING, WE GET THAT X = 1/5 SINCER, THIS METHOD WILL "NOT¢WORK FOR EVEN SLIGHTLY MORE COMPLICATEÄEQUATIONS. FOR EXAMPLEº SOLVE: 5X = 1 (MEANS FIVE TIMES ØEQUALS ONE). NOTE: THE NUMBER FIVE IÓCALLED THE "COEFFICIENT" OF X. WE CAÎNOT SOLVE THIS EQUATION WITH THE ADDI­TIVE RULE ALONET X - 10 = X (-10)¬SO ADD +10 TO BOTH SIDES. THEN X - 10 + 10 = X + (-10) + 10 = X AND 15 + 1°= 25. SO X = 25.   LET'S TRY AGAIN.  (PRESS RETURN TO GO ON)  €¼“ SO FAR WE HAVE USED THE ADDITIVÅRULE TO SOLVE SIMPLE ALGEBRAIC EQUA­TIONS. HOWEV ANSWER: ADD -6 TO BOTH SIDES, THEÎ X + 6 + (-6) = 7 + (-6) = 1    SO X = 1   LET'S TRY ANOTHER.   (PRESS RETURN TO GO ON©€Ú“ SORRY ,  HERE'S A SIMILAR EXAMPLE.   SOLVE: X - 10 = 15    X = ?    REMEMBER THARETURN TO GO ON©€Ê” NOT QUITE ®LET'S GO BACK AND REVIEW THIS® (PRESS RETURN TO GO ON) €¬”  8 - (-2) = ¿  1² -1° 8 + ² NONE OF THE ABOVÅ„<‘<‘\”<‘f” NO ®HERE'S A SIMILAR EXAMPLE:    SOLVE: X + 6 = 7 X = ?   PRESS RETURN TO GO ON) €j• EXCELLENT ¡5 + (-2) = 3® (PRESS RETURN TO GO ON) €ò” VERY GOOD ¡LET'S SEE WHY.   PRESS (A) TO SEE ILLUSTRATIOÎ PRESS (B) TO GO OÎ‚Þ”Ô”è” ¬THAT ISN'T CORRECT. PLEASE TRY AGAIN® (PRESS BOTH TRUE ANÄFALSE SIMULTANEOUSLY® (PRESS RETURN TO GO ON) €°• SORRY ®LET'S READ ABOUT POSITIVE AND NEGATIVÅNUMBERS AGAIN.  (PRESS RETURN TO GO ON©€°• INCORRECT ®LET'S REVIEW HOW TO ADD POSITIVE ANÄNEGATIVE NUMBERS.  ( ON) €Ê” EXCELLENT ¡THE COMMUTATIVE LAW STATES THAT WE CAÎADD POSITIVE AND NEGATIVE NUMBERS IN ANY ORDER® (PRESS RETURN TO GO ON) €¬” NO ®PLEASE TRY AGAIN.  (PRESS RETURN TO GO ON©€Ø• RIGHT ¡AN EQUATION CAN NOT BE OF A QUANTITY, RIGHÔ PRESENCE OF A QUANTITY, RIGHÔ„~•~•~•t•ˆ• ADDITION SATISFIES THE "COMMUTA­TIVE LAW". THE COMMUTATIVE LAW STATEÓTHAT WE CAN ADD TWO NUMBERS IN ANÙORDER. FOR EXAMPLEº  2 + 3 = 3 + 2 ; (-3) + 8 = 8 + (-3© (PRESS RETURN TO GODEA® (PRESS RETURN TO GO ON©€þ’ SOLVE: 2X - 4 = 0 X = ¿  ² ´ -² NONE OF THE ABOVÅ„^’T’T’T’h’ POSITIVE NUMBERS REPRESENT THÅ---?--- AND ARE TO THE ---?--- OF ZERO® ABSENCE OF A QUANTITY, LEFÔ PRESENCE OF A QUANTITY, LEFÔ ABSENCE, IF X + 3 = 1° THEN X + 3 + (-3) = X + 0 = Ø AND 10 + (-3) = 7 SO X = 7® (PRESS RETURN TO GO ON©€ø“ SOLVE: 10X = 2° X = ¿  ² ± 1/² NONE OF THE ABOVÅ„0“&“&“&“:“ EXACTLY ¡YOU SEEM TO BE GETTING THE I SIDE OF THE EQUATION. IN THÅ EXAMPLE X + 3 = 10 WE ADDED -³ TO BOTH SIDES TO GET X BY ITSELF ON THE LEFT. ON THE RIGHT SIDÅ WE WILL HAVE: 10 + (-3) = ·   3) PUT IT TOGETHER AND WE'RE DONE. IN OTHERWORDS THE EQUATION CAN BE FALSÅ THE EQUATION CAN BE TRUÅ BOTH (A) AND (B© NONE OF THE ABOVńΕΕº•Î•Ø• WE NOW KNOW HOW TO ADD POSITIVÅ"AND" NEGATIVE NUMBERS® PRESS (A) TO SEE ILLUSTRATIOÎ PRESS (B) TO GO O΂.•$•B• 2) COLLECT THE TERMS ON THE OTHER-A NEGATIVE NUMBEÒ NONE OF THE ABOVÅ„Œ‘Œ‘‚‘Œ‘–‘ REVIE×  REAL WORLD PROBLEMS ARE USUALLY RE-FERRED TO ASº WORD PROBLEMS  BIG TROUBLÅ EQUATIONÓ NONE OF THE ABOVÅ„ ‘‘‘‘‘ WHICH OF THE FOLLOWING IS "NOT¢TRUE OF A SINGLE EQUATION¿HAT IS ALGEBRA¿ THE STUDY OF QUANTITIEÓ THE STUDY OF EQUATIONÓ THE STUDY OF NUMBERÓ NONE OF THE ABOVÅ„¾‘´‘¾‘¾‘È‘ REVIE×   SUBTRACTING A POSITIVE NUMBER IÓTHE SAME ASº ADDING A POSITIVE NUMBEÒ SUBTRACTING A NEGATIVE NUMBEÒ ADDING (PRESS RETURN TO GO ON©€"’ VERY GOOD ¡ (PRESS RETURN TO GO ON©€Ü‘ FANTASTIC ¡YOU HAVE COMPLETED THE CONCEPT AND QUIÚSECTION. NOW LET'S GO TO THE REVIE×QUESTIONS® (PRESS RETURN TO GO ON©€È‘ REVIE×  WN 2X + 3 = 25® THE FOURTH AND FINAL STEP IS TÏSOLVE THE EQUATION AND FIND THE UNKNOWN.IN OUR EXAMPLE WE GET X = 11, AND SÏJANE IS 11 YEARS OLD® (PRESS RETURN TO GO ON©€’ SORRY ®LET'S GO BACK AND READ ABOUT SOLVINÇWORD PROBLEMS AGAIN.  NOW¬THEN SHE WOULD BE 3 YEARS YOUNGER THAÎMARY, WHO IS 25® (PRESS RETURN TO GO ON©€’ THE THIRD STEP IS TO WRITE DOWN AÎEQUATION ASSOCIATED WITH THE WORD PROB­LEM BASED ON WHAT YOU KNOW. IN OUR EX­AMPLE, LETTING X = JANE'S AGE WE GET THEEQUATIOST STEP IN SOLVING A WORÄPROBLEM IS TO DETERMINE THE "UNKNOWN"®THAT IS THE NUMBER YOU ARE TRYING TÏFIND. IN OUR EXAMPLE THE UNKNOWN IÓJANE'S AGE® THE SECOND STEP IS DETERMINE WHAÔYOU KNOW. IN OUR EXAMPLE WE KNOW THAÔIF JANE WAS TWICE AS OLD AS SHE IS ¡ (PRESS RETURN TO GO ON©€,’ YOU NOW KNOW HOW TO SOLVE THÅSIMPLEST TYPE OF ALGEBRAIC EQUATIONÓ®BUT WHY WOULD ONE WANT TO SOLVE SUCÈPROBLEMS? HOW DO THESE PROBLEMS OCCUÒIN THE REAL WORLD¿ (PRESS RETURN TO GO ON) €"’ THE FIRN TO GO ON©€|’ YES ¡YOU'RE DOING GREAT¡ (PRESS RETURN TO GO ON©€h’ SOLVE: 8X + 7 = 23 X = ¿  30/¸ 1¶ ² NONE OF THE ABOVÅ„@’@’6’@’J’ TOO BAD ®PLEASE TRY THE PROBLEM AGAIN® (PRESS RETURN TO GO ON©€J’ SUPERB= -6 + (-4) AND SO 5X = -10. NOW APPLY THE MULTIPLICATIVE RULE: (1/5) * 5X = (1/5) * (-10©SO X = -2 SINCE (1/5) * (-10) = -2®PLEASE TRY THE PROBLEM AGAIN® (PRESS RETURN TO GO ON©€š’ PRECISELY ¡THAT WAS A TOUGH ONE¡ (PRESS RETUREAT ¡ (PRESS RETURN TO GO ON©€š’ SOLVE: 10X + 3 = -27 X = ¿  24/1° ³ -³ NONE OF THE ABOVÅ„’’†’’š’ SORRY ®HERE IS A SIMILAR EXAMPLEº  SOLVE: 5X + 4 = -6 X = ¿  FIRST USE THE ADDITIVE RULEº5X + 4 + (-4) E'S A GENERAL METHOD FOR SOLVINGEQUATIONS LIKE 5X + 3 = 7® "FIRST" APPLY THE "ADDITIVE RULE"º 5X + 3 + (-3) = 7 + (-3©THEN 5X = 4® "SECOND" APPLY THE "MULTIPLCATIVÅRULE": (1/5) * 5X = (1/5) * ´SO X = 4/µ (PRESS RETURN TO GO ON©€®’ GRIMES0 EQUALS 0® (PRESS RETURN TO GO ON©€à’ VERY GOOD ¡ (PRESS RETURN TO GO ON©€à’ SOLVE: 10X = 100 X = ¿  1° 10° 1/1° NONE OF THE ABOVÅ„Ö’Ì’Ì’Ì’à’ PERFECT ¡10 * 10 = 100® (PRESS RETURN TO GO ON©€Â’ HER X = ¿  3/² 2/³ ° NONE OF THE ABOVÅ„“““““ SOLVE: 8X = 0 X = ¿  ± -± ° NONE OF THE ABOVÅ„ô’ô’ê’ô’þ’ NO ®THE CORRECT ANSWER WAS (C) 0. THIS IÓEASY TO SEE SINCE ANY NUMBER MULTIPLIEÄBY 0 EQUALS 0, AND IN PARTICULAR 8 T X = ¿  ¶ ± ° NONE OF THE ABOVÅ„N“D“N“N“X“ GREAT ¡IF 6X = 6, THEN X MUST BE ONE® (PRESS RETURN TO GO ON©€:“ EXCELLENT ¡IF 10X = 20, THEN X = 2. LET'S TRÙANOTHER® (PRESS RETURN TO GO ON©€“ SOLVE: 3X = ²   BOTH SIDES OF AÎEQUATION BY THE SAME NUMBER, AND THÅEQUATION WILL STILL BE TRUE® FOR EXAMPLE: IF 5X = 1, THÅ(1/5) * 5 = (1/5) * 1 (NOTICE WE WILÌUSE THE ASTERISK, *, TO MEAN MULTIPLICA-TION)® (PRESS RETURN TO GO ON©€l“ SOLVE: 6X = ¶ AND SO X = 15 + (-17) = -² PLEASE TRY AGAIN® (PRESS RETURN TO GO ON©€ž“ IN ORDER TO SOLVE THE EQUATIOÎ5X = 1, WE NEED TO USE THE "MULTIPLICA­TIVE RULE FOR EQUATIONS". THIS ROLE IÓSIMILAR TO THE ADDITIVE RULE, BUT STATESTHAT WE CAN "MULTIPLY" THAT ISN'T CORRECT ®HERE'S A SIMILAR EXAMPLEº   SOLVE: 15 - X = 1· ADD X TO BOTH SIDES TO GEÔ 15 - X + X = 17 + Ø BUT THEN 15 = 17 + Ø TRY ADDING -17 TO BOTH SIDES ANÄ WE HAVE THATº 15 + (-17) = 17 + X + (-17© 5 = ·  X = ¿  · -1² 1² NONE OF THE ABOVÅ„²“²“¨“²“¼“ THAT'S RIGHT ¡LET'S GO TO ANOTHER EQUATION® (PRESS RETURN TO GO ON©€ž“  SOLVE: 10 - X = 2°  X = ¿   ° 1° -1° NONE OF THE ABOVÅ„”“”“Š“”“ž“ NONE OF THE ABOVńГГƓГړ TOO BAD ®LET'S GO BACK AND READ ABOUT HOW TO USÅTHE ADDITIVE RULE TO SOLVE EQUATIONS  AGAIN® (PRESS RETURN TO GO ON©€ ” FANTASTIC ¡LET'S TRY ANOTHER® (PRESS RETURN TO GO ON©€¼“ SOLVE: X - X = ¿  µ -µ 1° NONE OF THE ABOVÅ„î“ä“ä“ä“ø“ THAT'S RIGHT ¡X = 5 BECAUSE X + 5 + (-5) = 10 + (-5)¬THUS X + 0 = 10 + (-5) = X = 5.  (PRESS RETURN TO GO ON©€Ú“   SOLVE: X + 10 = 2µ  X = ¿  1° -1µ 1µN JUST EQUALS X®  FOR EXAMPLE: X + 3 = 10 ADD -3 TÏBOTH SIDES SINCE   X + 3 + (-3) = X + 0 = Ø AND WE WILL GET X BY ITSELF ON THE LEFÔSIDE OF THE EQUATION® (PRESS RETURN TO GO ON©€” SOLVE: X + 5 = 1°  E®  FOR EXAMPLE IF X + 2 = ² THEN X + 2 + (-2) = 2 + (-2© (PRESS RETURN TO GO ON©€ ” WE CAN USE THE ADDITIVE RULE TÏSOLVE EQUATIONS AS FOLLOWSº 1) ADD A NUMBER TO BOTH SIDES OÆ AN EQUATION SO THAT ONE SIDÅ OF THE EQUATIO + 7 = ´ (PRESS RETURN TO GO ON©€” THERE IS A GENERAL METHOD FOÒSOLVING SIMPLE EQUATIONS. IT USES THÅ"ADDITIVE RULE FOR EQUATIONS". THÅADDITIVE RULE STATES THAT WE CAN ADD ÁNUMBER TO BOTH SIDES OF AN EQUATION, ANDTHE EQUATION WILL STILL BE TRUVOLVING INDETERMINATESº  X + 3 = ·  A + (-2) = -¶  B - 3 = ° (PRESS RETURN TO GO ON©€ ” THE MAIN GOAL OF ALGEBRA IS TÏ"SOLVE" EQUATIONS INVOLVING INDETERMI­NATES. FOR EXAMPLEº  X + 7 = 4 WHAT IS X¿  ANSWER: X = -3 SINCE (-3)NTITY (EITHER POSI­TIVE OR NEGATIVE) IS CALLED AN "INDETER-MINATE"® "LETTERS" (A, B, C...) ARE NORMALLYUSED TO REPRESENT INDETERMINATES® THE MOST COMMONLY USED LETTER IÓ"X"® (PRESS RETURN TO GO ON©€R” HERE ARE SOME EXAMPLES OF EQUATIONSIN THE ABOVÅ„>”4”4”4”H” SORRY ®REMEMBER THE COMMUTATIVE LAW. (-4) + 5 = 5 + (-4) = 5 - 4. THE RIGHT ANSWEÒWAS (A) 5 - 4® (PRESS RETURN TO GO ON©€*” VERY GOOD ¡(-4) + 5 = 5 - 4® (PRESS RETURN TO GO ON©€*” AN UNKNOWN QUA”„”Ž”Ž”˜” MARVELOUS ¡(-10) - 2 = -12® (PRESS RETURN TO GO ON©€f” CORRECT ¡SUBTRACTING A NEGATIVE NUMBER IS THÅSAME AS ADDING A POSITIVE NUMBER® (PRESS RETURN TO GO ON©€H”  (-4) + 5 = ¿  5 - ´ 5 + ´ -¹ NONE OFNG A POSITIVE NUMBER. FOR EXAMPLEº  4 - (-3) = 4 + 3 = ·  (-4) - (-4) = (-4) + 4 = °  (-10) + 4 = (-10) - (-4) = -¶  (PRESS TO GO ON©€˜” NOW YOU TRY AN EQUATION®  6 - 2 = ¿ 6 + ² 6 + (-2© ³ NONE OF THE ABOVÅ„Ž” SUBTRACTING A "POSITIVE NUMBER" IÓTHE SAME AS ADDING A NEGATIVE NUMBER®FOR EXAMPLEº 4+ (-3) = 4 - 3 = ±  10 - 7 = 10 + (-7) = ³  (-5) - 5 = (-5) + (-5) = -1° (PRESS TO GO ON©€¢” SUBTRACTING A "NEGATIVE NUMBER" IÓTHE SAME AS ADDI) TO SEE ILLUSTRATIOÎ PRESS (B) TO GO O΂ü”B•• THE COMMUTATIVE LAW STATES THATº WE CAN ADD POSITIVE NUMBERÓ WE CAN ADD NEGATIVE NUMBERÓ WE CAN ADD POSITIVE AND NEGATIVÅ NUMBERÓ WE CAN ADD POSITIVE AND NEGATIVÅ NUMBERS IN ANY ORDEÒ„À”À”À”¶”ÊRRESPONDS TO A MOVEMENT TO THE "LEFT¢ON THE NUMBER LINE® PRESS (A) TO SEE ILLUSTRATIOÎ PRESS (B) TO GO O΂L••V• IF YOU ADD A POSITIVE NUMBER AND ÁNEGATIVE NUMBER TOGETHER YOU GET A POSI-TIVE "OR" A NEGATIVE NUMBER DEPENDING ONTHE EQUATION® PRESS (A THIS IS BECAUSE ADDINGA POSITIVE NUMBER CORRESPONDS TO A MOVE-MENT TO THE RIGHT ON THE NUMBER LINE® PRESS (A) TO SEE ILLUSTRATIOÎ PRESS (B) TO GO O΂`•V•j• 5 + (-2) = ¿ µ ² ³ NONE OF THE ABOVÅ„••••$• SIMILARLY ADDING A NEGATIVE NUMBEÒCOVE NUMBERS ARE TO THÅRIGHT OF "ZERO" (0), WHILE THE NEGATIVÅNUMBERS ARE TO THE LEFT OF "ZERO" (0)® (PRESS RETURN TO GO ON©€ˆ• HOW CAN WE ADD A POSITIVE AND ÁNEGATIVE NUMBER? REMEMBER WHEN WE ADÄTWO POSITIVE NUMBERS WE GET ANOTHEÒPOSITIVE NUMBER.p ¹1012,0#(ºç(4);"BRUNTURBO"A2ºç(4);"BLOADRUNNER,A$880"^<ºç(4);"BLOADQUIZ,A$2000"hFŒ2176nP€É­ FOR EXAMPLE: THE EQUATIOÎ2 + 2 = 4 IS TRUE, WHILE THE EQUATIOÎ1 = 0 IS FALSE® (PRESS RETURN TO GO ON©€Ø• WE CAN THINK OF POSITIVE AND NEGA­TIVE NUMBERS AS POINTS ON A LINE® PRESS (A) TO SEE ILLUSTRATIOÎ PRESS (B) TO GO O΂œ•’•¦• THE POSITIvìfj– ALGEBRA IS THE STUDY OF "EQUA­TIONS". AN EQUATION STATES THAT TWÏNUMERICAL "QUANTITIES" ARE THE SAME®FOR EXAMPLEº  2 + 2 = ´  0 + 10 = 1° (PRESS RETURN TO GO ON©€ì• AN EQUATION IS EITHER "TRUE" OÒ"FALSE".                                                           þz#ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ    ÿLM¥ë8é …ë°Æì`Á$ * ±ëÉÿð Á·®ÐÉà                       " Á¶®ÐÉà                       " Á¸®ÐÉà                       "Á¹®ÐÉà                       "ÿÁ²®¶®ÐÉà                     "ÿÁ³±®ÐÉà                     !"ÿÁ²·®ÐÉà                     "" "ÒÕÎÎÅÒ                         ÈÅÌÌÏ                         ÑÕÉÚ                          x®ÔÅÍЮ                        "ÔÕÒÂÏ                         Á³®ÐÉà                       "Á±¸®ÐÉà                      "€€€€€€€€€ 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WE WILL GET X BY ITSELF ON THE LEFÔSIDE OF THE EQUATION® (PRESS RETURN TO GO ON©€” SOLVE: X + 5 = 1°  E®  FOR EXAMPLE IF X + 2 = ² THEN X + 2 + (-2) = 2 + (-2© (PRESS RETURN TO GO ON©€ ” WE CAN USE THE ADDITIVE RULE TÏSOLVE EQUATIONS AS FOLLOWSº 1) ADD A NUMBER TO BOTH SIDES OÆ AN EQUATION SO THAT ONE SIDÅ OF THE EQUATIO + 7 = ´ (PRESS RETURN TO GO ON©€” THERE IS A GENERAL METHOD FOÒSOLVING SIMPLE EQUATIONS. IT USES THÅ"ADDITIVE RULE FOR EQUATIONS". THÅADDITIVE RULE STATES THAT WE CAN ADD ÁNUMBER TO BOTH SIDES OF AN EQUATION, ANDTHE EQUATION WILL STILL BE TRUVOLVING INDETERMINATESº  X + 3 = ·  A + (-2) = -¶  B - 3 = ° (PRESS RETURN TO GO ON©€ ” THE MAIN GOAL OF ALGEBRA IS TÏ"SOLVE" EQUATIONS INVOLVING INDETERMI­NATES. FOR EXAMPLEº  X + 7 = 4 WHAT IS X¿  ANSWER: X = -3 SINCE (-3)NTITY (EITHER POSI­TIVE OR NEGATIVE) IS CALLED AN "INDETER-MINATE"® "LETTERS" (A, B, C...) ARE NORMALLYUSED TO REPRESENT INDETERMINATES® THE MOST COMMONLY USED LETTER IÓ"X"® (PRESS RETURN TO GO ON©€R” HERE ARE SOME EXAMPLES OF EQUATIONSIN THE ABOVÅ„>”4”4”4”H” SORRY ®REMEMBER THE COMMUTATIVE LAW. (-4) + 5 = 5 + (-4) = 5 - 4. THE RIGHT ANSWEÒWAS (A) 5 - 4® (PRESS RETURN TO GO ON©€*” VERY GOOD ¡(-4) + 5 = 5 - 4® (PRESS RETURN TO GO ON©€*” AN UNKNOWN QUA”„”Ž”Ž”˜” MARVELOUS ¡(-10) - 2 = -12® (PRESS RETURN TO GO ON©€f” CORRECT ¡SUBTRACTING A NEGATIVE NUMBER IS THÅSAME AS ADDING A POSITIVE NUMBER® (PRESS RETURN TO GO ON©€H”  (-4) + 5 = ¿  5 - ´ 5 + ´ -¹ NONE OFNG A POSITIVE NUMBER. FOR EXAMPLEº  4 - (-3) = 4 + 3 = ·  (-4) - (-4) = (-4) + 4 = °  (-10) + 4 = (-10) - (-4) = -¶  (PRESS TO GO ON©€˜” NOW YOU TRY AN EQUATION®  6 - 2 = ¿ 6 + ² 6 + (-2© ³ NONE OF THE ABOVÅ„Ž” SUBTRACTING A "POSITIVE NUMBER" IÓTHE SAME AS ADDING A NEGATIVE NUMBER®FOR EXAMPLEº 4+ (-3) = 4 - 3 = ±  10 - 7 = 10 + (-7) = ³  (-5) - 5 = (-5) + (-5) = -1° (PRESS TO GO ON©€¢” SUBTRACTING A "NEGATIVE NUMBER" IÓTHE SAME AS ADDI) TO SEE ILLUSTRATIOÎ PRESS (B) TO GO O΂ü”B•• THE COMMUTATIVE LAW STATES THATº WE CAN ADD POSITIVE NUMBERÓ WE CAN ADD NEGATIVE NUMBERÓ WE CAN ADD POSITIVE AND NEGATIVÅ NUMBERÓ WE CAN ADD POSITIVE AND NEGATIVÅ NUMBERS IN ANY ORDEÒ„À”À”À”¶”ÊRRESPONDS TO A MOVEMENT TO THE "LEFT¢ON THE NUMBER LINE® PRESS (A) TO SEE ILLUSTRATIOÎ PRESS (B) TO GO O΂L••V• IF YOU ADD A POSITIVE NUMBER AND ÁNEGATIVE NUMBER TOGETHER YOU GET A POSI-TIVE "OR" A NEGATIVE NUMBER DEPENDING ONTHE EQUATION® PRESS (A THIS IS BECAUSE ADDINGA POSITIVE NUMBER CORRESPONDS TO A MOVE-MENT TO THE RIGHT ON THE NUMBER LINE® PRESS (A) TO SEE ILLUSTRATIOÎ PRESS (B) TO GO O΂`•V•j• 5 + (-2) = ¿ µ ² ³ NONE OF THE ABOVÅ„••••$• SIMILARLY ADDING A NEGATIVE NUMBEÒCOVE NUMBERS ARE TO THÅRIGHT OF "ZERO" (0), WHILE THE NEGATIVÅNUMBERS ARE TO THE LEFT OF "ZERO" (0)® (PRESS RETURN TO GO ON©€ˆ• HOW CAN WE ADD A POSITIVE AND ÁNEGATIVE NUMBER? REMEMBER WHEN WE ADÄTWO POSITIVE NUMBERS WE GET ANOTHEÒPOSITIVE NUMBER. FOR EXAMPLE: THE EQUATIOÎ2 + 2 = 4 IS TRUE, WHILE THE EQUATIOÎ1 = 0 IS FALSE® (PRESS RETURN TO GO ON©€Ø• WE CAN THINK OF POSITIVE AND NEGA­TIVE NUMBERS AS POINTS ON A LINE® PRESS (A) TO SEE ILLUSTRATIOÎ PRESS (B) TO GO O΂œ•’•¦• THE POSITIÑ ìfj– ALGEBRA IS THE STUDY OF "EQUA­TIONS". AN EQUATION STATES THAT TWÏNUMERICAL "QUANTITIES" ARE THE SAME®FOR EXAMPLEº  2 + 2 = ´  0 + 10 = 1° (PRESS RETURN TO GO ON©€ì• AN EQUATION IS EITHER "TRUE" OÒ"FALSE".            NOW¬THEN SHE WOULD BE 3 YEARS YOUNGER THAÎMARY, WHO IS 25® (PRESS RETURN TO GO ON©€’ THE THIRD STEP IS TO WRITE DOWN AÎEQUATION ASSOCIATED WITH THE WORD PROB­LEM BASED ON WHAT YOU KNOW. IN OUR EX­AMPLE, LETTING X = JANE'S AGE WE GET THEEQUATIOST STEP IN SOLVING A WORÄPROBLEM IS TO DETERMINE THE "UNKNOWN"®THAT IS THE NUMBER YOU ARE TRYING TÏFIND. IN OUR EXAMPLE THE UNKNOWN IÓJANE'S AGE® THE SECOND STEP IS DETERMINE WHAÔYOU KNOW. IN OUR EXAMPLE WE KNOW THAÔIF JANE WAS TWICE AS OLD AS SHE IS ¡ (PRESS RETURN TO GO ON©€,’ YOU NOW KNOW HOW TO SOLVE THÅSIMPLEST TYPE OF ALGEBRAIC EQUATIONÓ®BUT WHY WOULD ONE WANT TO SOLVE SUCÈPROBLEMS? HOW DO THESE PROBLEMS OCCUÒIN THE REAL WORLD¿ (PRESS RETURN TO GO ON) €"’ THE FIRN TO GO ON©€|’ YES ¡YOU'RE DOING GREAT¡ (PRESS RETURN TO GO ON©€h’ SOLVE: 8X + 7 = 23 X = ¿  30/¸ 1¶ ² NONE OF THE ABOVÅ„@’@’6’@’J’ TOO BAD ®PLEASE TRY THE PROBLEM AGAIN® (PRESS RETURN TO GO ON©€J’ SUPERB= -6 + (-4) AND SO 5X = -10. NOW APPLY THE MULTIPLICATIVE RULE: (1/5) * 5X = (1/5) * (-10©SO X = -2 SINCE (1/5) * (-10) = -2®PLEASE TRY THE PROBLEM AGAIN® (PRESS RETURN TO GO ON©€š’ PRECISELY ¡THAT WAS A TOUGH ONE¡ (PRESS RETUREAT ¡ (PRESS RETURN TO GO ON©€š’ SOLVE: 10X + 3 = -27 X = ¿  24/1° ³ -³ NONE OF THE ABOVÅ„’’†’’š’ SORRY ®HERE IS A SIMILAR EXAMPLEº  SOLVE: 5X + 4 = -6 X = ¿  FIRST USE THE ADDITIVE RULEº5X + 4 + (-4) E'S A GENERAL METHOD FOR SOLVINGEQUATIONS LIKE 5X + 3 = 7® "FIRST" APPLY THE "ADDITIVE RULE"º 5X + 3 + (-3) = 7 + (-3©THEN 5X = 4® "SECOND" APPLY THE "MULTIPLCATIVÅRULE": (1/5) * 5X = (1/5) * ´SO X = 4/µ (PRESS RETURN TO GO ON©€®’ GRIMES0 EQUALS 0® (PRESS RETURN TO GO ON©€à’ VERY GOOD ¡ (PRESS RETURN TO GO ON©€à’ SOLVE: 10X = 100 X = ¿  1° 10° 1/1° NONE OF THE ABOVÅ„Ö’Ì’Ì’Ì’à’ PERFECT ¡10 * 10 = 100® (PRESS RETURN TO GO ON©€Â’ HER X = ¿  3/² 2/³ ° NONE OF THE ABOVÅ„“““““ SOLVE: 8X = 0 X = ¿  ± -± ° NONE OF THE ABOVÅ„ô’ô’ê’ô’þ’ NO ®THE CORRECT ANSWER WAS (C) 0. THIS IÓEASY TO SEE SINCE ANY NUMBER MULTIPLIEÄBY 0 EQUALS 0, AND IN PARTICULAR 8 T X = ¿  ¶ ± ° NONE OF THE ABOVÅ„N“D“N“N“X“ GREAT ¡IF 6X = 6, THEN X MUST BE ONE® (PRESS RETURN TO GO ON©€:“ EXCELLENT ¡IF 10X = 20, THEN X = 2. LET'S TRÙANOTHER® (PRESS RETURN TO GO ON©€“ SOLVE: 3X = ²   BOTH SIDES OF AÎEQUATION BY THE SAME NUMBER, AND THÅEQUATION WILL STILL BE TRUE® FOR EXAMPLE: IF 5X = 1, THÅ(1/5) * 5 = (1/5) * 1 (NOTICE WE WILÌUSE THE ASTERISK, *, TO MEAN MULTIPLICA-TION)® (PRESS RETURN TO GO ON©€l“ SOLVE: 6X = ¶ AND SO X = 15 + (-17) = -² PLEASE TRY AGAIN® (PRESS RETURN TO GO ON©€ž“ IN ORDER TO SOLVE THE EQUATIOÎ5X = 1, WE NEED TO USE THE "MULTIPLICA­TIVE RULE FOR EQUATIONS". THIS ROLE IÓSIMILAR TO THE ADDITIVE RULE, BUT STATESTHAT WE CAN "MULTIPLY" THAT ISN'T CORRECT ®HERE'S A SIMILAR EXAMPLEº   SOLVE: 15 - X = 1· ADD X TO BOTH SIDES TO GEÔ 15 - X + X = 17 + Ø BUT THEN 15 = 17 + Ø TRY ADDING -17 TO BOTH SIDES ANÄ WE HAVE THATº 15 + (-17) = 17 + X + (-17© 5 = ·  X = ¿  · -1² 1² NONE OF THE ABOVÅ„²“²“¨“²“¼“ THAT'S RIGHT ¡LET'S GO TO ANOTHER EQUATION® (PRESS RETURN TO GO ON©€ž“  SOLVE: 10 - X = 2°  X = ¿   ° 1° -1° NONE OF THE ABOVÅ„”“”“Š“”“ž“ NONE OF THE ABOVńГГƓГړ TOO BAD ®LET'S GO BACK AND READ ABOUT HOW TO USÅTHE ADDITIVE RULE TO SOLVE EQUATIONS  AGAIN® (PRESS RETURN TO GO ON©€ ” FANTASTIC ¡LET'S TRY ANOTHER® (PRESS RETURN TO GO ON©€¼“ SOLVE: X - X = ¿  µ -µ 1° NONE OF THE ABOVÅ„î“ä“ä“ä“ø“ THAT'S RIGHT ¡X = 5 BECAUSE X + 5 + (-5) = 10 + (-5)¬THUS X + 0 = 10 + (-5) = X = 5.  (PRESS RETURN TO GO ON©€Ú“   SOLVE: X + 10 = 2µ  X = ¿  1° -1µ 1µ€€¾€€€¾€€€€Œ€€€€€€€€‚€€€€‚€€€€‚€€€À€€€€€‚€€„€€‚€€€€ „€€„€à‡€€„€À€€€€à„€€€À€€€„€à‡€€„€ „€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€ƒ€€Àñ€˜€€€Œ€€€Àƒ€àÿ€€€€Àƒ€€€¸€€€ðÀƒ€€€Œ€€ðÿƒ€€€€€€€ðÿƒ€€‚’‚¢¢€’‚ˆ¢’²€ˆ¢€¢¢²ˆˆ²¢‚ˆ€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€Ÿ€€þ¿€€¸€üŸð€‡€€€€€€üŸÀƒ€€€€€€€€€€€€€€‚€€€€€€€€€€€€€€€€€€€€€€€€€ €€€€€€€€€€€€€€‚€€€€€‚€€€€€‚€€€À€€€€€‚€€„€€‚€‚€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€‚€€€€‚€€€€€€€€À€€€€€€‚€€„€€‚€€‚àïƒà„ூàƒàƒà€ …€€€€ƒ€€ƒ€ ‚€€„€àƒ€€€€€þ¿€€€€€€€‚€€ü€€€€€€€€€ð€€€€€€†€€€ó€Œ€€€Œ€€ðÀƒ€€€€€€€ðÀƒ€€€Œ€€€ð€†€€€Œ€€ðÿƒ€€€€€€€ðÿƒ€€ˆ‚Š‚  €Š‚ˆ¢Šº€ˆ¢€‚¢ºˆˆº¢‚€€€€€€€€€€€€€€€€€€€€€€€€€¼€€À€€¸€€€ðŸ€‡€€üŸ€€€€Àƒ€€€€€€€€€€€€€€€€€€‚œ‚€¾¢Üª‚œ¼˜€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€‚€€‚‚À€‚€„€€‚€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€Àƒ€Àƒ€€‚€Àƒ€à‡€à‡€Àƒ€à‡€à‡€Àƒ€€‚€Àƒ€Àƒ€€€€€€€€€€€€€€€€€€„€‚‚€…À€À‚‚€„€‚€ˆ€ €€àƒ€À‚€€„€€„€ € †€ €€„€€„€À‚€àƒ€ €€€€€€þ¿€€À€€€‡€€þŸ¸€€€€€€€€€ð€€€€€€Œ€€€ö€†€€€Œ€€ðჀ€€€€€€ðჀ€€†€€€ð€Œ€€€Œ€€ð€€€€€€€€€ð€€€€ˆžžžœœ€žžˆ¢žª€ˆ¢€‚¢ªˆˆª¢ž€€€€€€€€€€€€€€€€€€€€€€€€€€¸€€À€€°€€€€°€ƒ€€üŸ€€€€Àƒ€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€‚¢‚€‚”¢ªžˆ‚˜€€€€€€€€€€€€€€€€€€€€P€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€ „€ „€€‚€ „€ €€€€ „€€€ €€ „€€‚€ „€ „€€€€€€€€€€€€€€€€€€ÕªÕªÕªÕªÕªÕªÕªÕª……ÕªÕªÕªÕªժժժժժժ€À€€ €€€ƒ€ „€ „€À€ „€À€ „€ „€€ƒ€ €€À€€€€€€þ¿€€À€€À€€€¸à€€€€€€€€€ð€€€€€€¸€€€üÀƒ€€€Œ€€àÿ€€€€€€€àÿ€€€ƒ€€€ð€˜€€€Œ€€ð€€àÿ€€€ð€€€ˆ¢¢‚‚‚€¢‚ˆ¢¢ª€ˆ¢€‚¢ªˆˆª¢‚€ N 2X + 3 = 25® THE FOURTH AND FINAL STEP IS TÏSOLVE THE EQUATION AND FIND THE UNKNOWN.IN OUR EXAMPLE WE GET X = 11, AND SÏJANE IS 11 YEARS OLD® (PRESS RETURN TO GO ON©€’ SORRY ®LET'S GO BACK AND READ ABOUT SOLVINÇWORD PROBLEMS AGAIN.  Àƒàÿñÿƒ¸€€ðÿƒ€Àÿ€€€€€€àÿÁÿ€€€ žž¾œœ€ž¾¾¢ž¢€¾œ€œœ¢¾œ¢¢¾„€€ €€àƒ€À‚€€„€€„€ € †€ €€„€€„€À‚€àƒ€ €€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€üŸ€€À€€€‡€€þŸ¸€€€€€€€€€àƒ€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€ž€€€À€€°€€€€°€ƒ€€üŸ€€€€Àƒ€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€„€€€‚€À‚€€‚€À‚€€‚€€À€€‚€„‚€ˆ€€€€€€€€€€€€€€€€€€€€€€€€€€€ €€€€€€€€€‚‚€‚À€€‚€„‚€€€€‚œš€‚¢œ¿žˆœ˜€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€Àƒ€Àƒ€€‚€Àƒ€à‡€à‡€Àƒ€à‡€à‡€Àƒ€€‚€Àƒ€Àƒ€Àƒàÿ±ð€¸€€ðÿƒ€€ø€€€€€àÿø€€€À€€ €€€ƒ€ „€ „€À€ „€À€ „€ „€€ƒ€ €€À€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€ø€€€€€€€‚€€ü€€€€€€€€€Àƒ€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€üƒ€€À€€¸€€€€¸€‡€€üŸ€€€€Àƒ€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€ÕªÕªÕªÕªժÕªժժժժժժժժժժժժ€€€€€€€€€€€€€€€€€€€€€€€€€€€ €€€€€€€€€€€€€€€€‚€€€‚€€€€ˆ€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€ „€ „€€‚€ „€ €€€€ „€€€ €€ „€€‚€ „€ „€€€€Àƒ€€àð€¸€€€Œ€€€àƒ€àÿ€€€€àƒ€€€€‡€à‡€€‚€Àƒ€Àƒ€€€Àƒ€€€Àƒ€Àƒ€€‚€à‡€€‡€€€€ð€€€€€€€ðÿƒ€€€€€€€ðÿƒ€€ ‚¢¾œœ€¢¾ˆœ¢¢€ˆœ€œœ¢ˆœ¢œ¾„€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€ø€€þ¿€€¸€üŸ€œ€‡€€€€€€üŸÀƒ€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€ U*U*U €€€€€€€€€€€„€€€‚€À‚€€‚€À‚€€‚€€€À€€€€€‚€€„€€‚€€ˆ€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€é &$ Š‘Èh‘ + `ð ¥ðñ ˜‘æï ´ ¡$ÿ0Õ s ÿ¥ðóÐë¥ï8éA…þ € ‘ÈL¡­ïÉð? ˆTHIS IS THE NAME OF THE PICTURE FILE USED FOR THIS FRAMÅL ý`` ˆWHICH FRAME?  s¦ð2 Ø ¢°b¥i…¥i… Xü “ Žý¢ ±ðDÉÿÐLÁ ÉL™­ïÉð-Éð)­ …­ …¢ K ¥ÅÐ¥Åð@  ±ð±þ ‘ ±þ ‘¥ÿ¦þ j B ` ; ¢ Õ   ±Éв S  ©‘È©‘© ¢‰ o  ­‹ ‘È­Œ ‘¥ÿ¦þ o B `­Ò… „±Hȱ…h…Ð¥ðy ±ð $Ðå©‘ ©‘© ‘È‘ ‘ ã„<…= ±